Chapter 6.3 Rigidity and Local Linear Approximation

In this section we discuss important results involving the derivative of a function. We will discuss how the derivative can help us determine the extreme values of a function on a closed and bounded interval, and how the derivative can help us control the outputs of a function. We also discuss what it means for a function to have an antiderivative. Finally, we will discuss L’Hoptials rule, which is a way to evaluate limits of a certain kind.

Extreme Values

The derivative can reveal a lot about a function’s extreme values.

Both of the above functions have either a local min or local max at $x=2$. If we draw a tangent line at $x=2$, we get the following graphs

The tangent line has a slope of zero, which would mean the derivative is zero. Is this always true? To answer the question, we need to explore derivatives more.

Left Derivative

For any real valued function $f$, if there is a real number $a$ that is less than $x_0$ so that $f$ is defined on $(a, x_0]$, then denote by $f^\prime_-(x_0)$ the limit \[f^\prime_-(x_0) = \lim_{h\to 0^-} \frac{f(x_0 + h) - f(x_0)}{h},\] whenever this limit exists.

The value $f^\prime_-(x_0)$ is the left derivative of $f$ at $x_0$.

Right Derivative

For any real valued function $f$, if there is a real number $b$ that is greater than $x_0$ so that $f$ is defined on $[x_0, b)$, then denote by $f^\prime_+(x_0)$ the limit \[f^\prime_+(x_0) = \lim_{h\to 0^+} \frac{f(x_0 + h) - f(x_0)}{h},\] whenever this limit exists.

The value $f^\prime_+(x_0)$ is the right derivative of $f$ at $x_0$.

Differentiable Theorem

The function $f$ is differentiable at $x_0$ if and only if both $f^\prime_-(x_0)$ and $f^\prime_+(x_0)$ exist and are equal.

Example 1

For each function $f$ and each real number $x_0$ given below, determine $f^\prime_-(x_0)$ and $f^\prime_+(x_0)$ respectively:

$f(x) = |x|$, $x_0 = 0$;
$f(x) = \begin{cases}x^2 & \text{if } x< 1\\2-x^2 & \text{if } x\ge 1\end{cases}$, $x_0 = 1$;
$f(x) = |(x-5)^3|$, $x_0 = 5$.

Determine which of the above functions are differentiable at $x_0$.

To find $f'_-(0),$ calculate $\displaystyle\lim_{h\to 0^-}\frac{f(0+h)-f(0)}{h}.$ Do it to get \[\displaystyle\lim_{h\to 0^-}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0^-}\frac{f(h)-f(0)}{h}=\lim_{h\to 0^-}\frac{-h-0}{h}=\lim_{h\to 0^-}\frac{-h}{h}=-1.\] So $f'_-(0)=-1.$

To find $f'_+(0),$ calculate $\displaystyle\lim_{h\to 0^+}\frac{f(0+h)-f(0)}{h}.$ Do it to get \[\displaystyle\lim_{h\to 0^+}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0^+}\frac{f(h)-f(0)}{h}=\lim_{h\to 0^+}\frac{h-0}{h}=\lim_{h\to 0^+}\frac{h}{h}=1.\] So $f'_+(0)=1$.

Since the left and right derivative at $0$ are not equal, $f(x)=|x|$ is not differentiable at $0$.

To find $f'_-(1),$ calculate $\displaystyle\lim_{h\to 0^-}\frac{f(1+h)-f(1)}{h}.$ Do it to get \[\displaystyle\lim_{h\to 0^-}\frac{f(1+h)-f(1)}{h}=\lim_{h\to 0^-}\frac{(1+h)^2-f(1)}{h}=\lim_{h\to 0^-}\frac{1+2h+h^2-1}{h}=\lim_{h\to 0^-}\frac{2h+h^2}{h}=\lim_{h\to 0^-}(2+h)=2.\] So $f'_-(1)=2.$

To find $f'_+(1),$ calculate $\displaystyle\lim_{h\to 0^+}\frac{f(1+h)-f(1)}{h}.$ Do it to get \[\displaystyle\lim_{h\to 0^+}\frac{f(1+h)-f(1)}{h}=\lim_{h\to 0^+}\frac{2-(1+h)^2-1}{h}=\lim_{h\to 0^+}\frac{2-1-2h-h^2-1}{h}=\lim_{h\to 0^+}\frac{-2h-h^2}{h}=\lim_{h\to 0^+}(-2-h)=-2.\] So $f'_+(1)=-2$.

Since the left and right derivative at $1$ are not equal, $f$ is not differentiable at $1$.

Note that \[f(x)=\begin{cases}-(x-5)^3&\text{ if } x <5\\ (x-5)^3 &\text{ if }x\geq 5.\end{cases}\] To find $f'_-(5),$ calculate $\displaystyle\lim_{h\to 0^-}\frac{f(5+h)-f(5)}{h}.$ Do it to get \[\displaystyle\lim_{h\to 0^-}\frac{f(5+h)-f(5)}{h}=\lim_{h\to 0^-}\frac{-(5+h-5)^3-0}{h}=\lim_{h\to 0^-}\frac{-h^3-0}{h}=\lim_{h\to 0^-}\frac{-h^3}{h}=\lim_{h\to 0^-}-h^2=0.\] So $f'_-(5)=0.$

To find $f'_+(5),$ calculate $\displaystyle\lim_{h\to 0^+}\frac{f(5+h)-f(5)}{h}.$ Do it to get \[\displaystyle\lim_{h\to 0^+}\frac{f(5+h)-f(5)}{h}=\lim_{h\to 0^+}\frac{(5+h-5)^3-0}{h}=\lim_{h\to 0^+}\frac{h^3-0}{h}=\lim_{h\to 0^+}\frac{h^3}{h}=\lim_{h\to 0^+}h^2=0.\] So $f'_+(5)=0$.

Since the left and right derivative at $5$ are equal, $f$ is differentiable at $0$.

Note that $|x|$ is not differentiable at $x=0$, but it still has a local minimum at $x=0$. A function may still have a local min or local max even if the derivative is undefined. However, the following theorem says that if the function is differentiable and has a local max or min at $x_0$, then it’s derivative must be zero.

Fermat’s Theorem

Fermat’s Theorem: For any real valued function $f$ that is defined on an open interval $I$, if

$x_0$ is in $I$,
$f$ has a local maximum or minimum at $x_0$,
and $f$ is differentiable at $x_0$,

then $f^\prime(x_0)$ is equal to $0$.

We will prove that theorem by assuming that $x_0$ is a local minimum. If $f$ has a local minimum at $x_0$, then $-f$ has a local maximum at $x_0$, and $-f(x_0)$ is equal to $0$ if and only if $f(x_0)$ is equal to $0$.

For any $f$ that is defined on $(a, x_0)$ and has a local maximum at $x_0$, if $h$ is negative, then \[\frac{f(x_0+h) - f(x_0)}{h} \ge 0\quad \text{and so}\quad f_-^\prime(x_0)\ge 0.\]

For any $f$ that is defined on $[x_0, b)$ and has a local maximum at $x_0$, if $h$ is positive, then \[\frac{f(x_0+h) - f(x_0)}{h} \le 0\quad \text{and so}\quad f_+^\prime(x_0)\le 0.\]

Since $f$ is differentiable at $x_0$, \[f^\prime(x_0) = f_-^\prime(x_0) = f_+^\prime(x_0)\quad \text{and so}\quad f^\prime(x_0) = 0.\]

For any $f$ that is defined on $I$ and that has a local minimum at $x_0$, $-f$ is also differentiable and $-f$ has a local maximum at $x_0$.

The linearity of differentiation implies that \[f^\prime(x_0) = (-(-f)))^\prime(x_0) = -(-f)^\prime(x_0) = -0 =0.\]

Rolle’s Theorem

Rolle’s Theorem For any real values $a$ and $b$ with $a$ less than $b$ and for any real valued function $f$ that is defined on an interval $[a,b]$, if

$f$ is differentiable on $(a,b)$,
$f$ is continuous on $[a,b]$,
and $f(a)$ is equal to $f(b)$,

then there is a $c$ in $(a,b)$ so that \[f^\prime(c) = 0.\]

Here is why the theorem is true.

Continuous functions attain their maximum and minimum values on closed and bounded intervals, and so $f$ attains its maximum value $M$ and its minimum value $m$ in $[a,b]$.

If $f$ attains both values $M$ and $m$ on the set $\{a,b\}$, then $m$ is equal to $M$, and so $f$ is the constant function.
If $f(a)$ is not equal to $m$ (or $M$), then $f$ attains $m$ (or $M$) in $(a,b)$. Since $f(a)$ and $f(b)$ are equal, there is a point $c$ in $(a,b)$ at which either \[f(c) = m \quad \text{or}\quad f(c) = M.\] Fermat’s theorem implies that \[f^\prime(c) = 0.\]

Fermat’s theorem is important for many reasons.

A practical use is for locating the points at which a differentiable function $f$ on a closed interval $[a,b]$ attains its maximum and minimum values.

Determine the zero set $Z$ of $f^\prime$;
The function $f$ must attain its maximum and minimum values on the set $Z\cup \{a, b\}$;
Compare the values that $f$ takes on in the set $Z\cup \{a, b\}$ to determine the maximum and minimum values of $f$ on $[a,b]$.

Note that if $f$ is continuous on $[a,b]$ but fails to be differentiable at finitely many points $Z^\prime$, then it is still possible to use Fermat’s theorem to locate the points at which $f$ attains its maximum and minimum values on $[a,b]$.

To do so, compare the values that $f$ takes on in the set $Z\cup Z^\prime\cup \{a,b\}$.

The set $Z\cup Z^\prime$ is the set of critical points of $f$ in $[a,b]$.

Example 2

Determine all points at which each function $f$ below can potentially attain a local maximum or a local minimum:

$f(x) = x^2 - 10x + 7$;
$f(x) = x{\rm e}^{-x^2}$.
$f(x) = |x(x-1)^2|$.

The derivative of $f$ is $f'(x)=2x-10$. Find where $f'(x)=0$ by solving \[0=2x-10.\] The solution is $x=5.$ It is possible this is where the local max or local min of the function is. In fact, this is exactly where it is since $f$ can be rewritten as $(x-5)^2-18$. Because the parabola opens up, the local minimum is at $(5,-18).$

The derivative of $f$ is $f'(x)=e^{-x^2}-2x^2e^{-x^2}$, Find where $f'(x)=0$ by solving \[0=e^{-x^2}-2x^2e^{-x^2}.\] Do the work to get \[\begin{align*}0&=e^{-x^2}-2x^2e^{-x^2}\\0&=e^{-x^2}(1-2x^2)\\\end{align*}\] implies that $0=e^{-x^2}$ or $0=1-2x^2$. The first equation has no solution: the exponential function is never zero. The second equation has solution $x=\pm \frac{1}{\sqrt{2}}$. So it is possible there is a local maximum or local minimum at $x=\frac{1}{\sqrt{2}}$ and $x=-\frac{1}{\sqrt{2}}$.

The function $f$ is actually a piecewise function \[f(x)=\begin{cases}-x(x-1)^2&\text{ if }x<0\\x(x-1)^2&\text{ if }x\ge 0\end{cases}.\] So when $x<0$, $f'(x)=-(x-1)^2-2x(x-1)=-3x^2+4x-1$. When $x>0$, $f'(x)=(x-1)^2+2x(x-1)=3x^2-4x+1$. Because $f'_{-}(0)=-1$ and $f'_{+}(0)=1$, the function is not differentiable at $x=0.$ There may or may not be a local maximum or local minimum. The derivative is equal to zero when the equation $-3x^2+4x-1=0$ on $(-\infty,0)$ and $3x^2-4x+1=0$ are $(0,\infty)$ are satisfied. The derivative is therefore zero at $x=\tfrac{1}{3}$ or $x=1$.

The conclusion is that the function may have a local maximum or local minimum at $x=0$, $x=\tfrac{1}{3}$ or $x=1$.

If we have a continuous function on a closed and bounded interval, then by the Extreme Value Theorem, it achieves its maximum and minimum on that interval. So finding the critical values can aid in identifying the maximum and minimum values. Try it out with this next example.

Example 3

For each of these choices of interval $[a,b]$ and function $f$, determine all points at which $f$ attains its maximum and minimum values on $[a,b]$:

$f(x) = x^2 - 10x + 7$ and $[-1, 6]$
$f(x) = x{\rm e}^{-x^2}$ and $[-3, 3]$
$f(x) = |x(x-1)^2|$ and $[-1, 2]$

Part a

Because $f$ is continuous on $[-1,6]$, it will achieve its maximum and minimum in $[-1,6]$. It can happen at the endpoints or at its critical values, which is $5.$ So, $f$ attains its maximum or minimum on the set $\{-1,5,6\}.$ Evaluate $f$ to see that

\[f(-1)=18,\quad f(5)=-18,\quad\text{and}\quad f(6)=-17.\]

So $f$ attains its minimum at $5$ and maximum at $18$.

Part b

Because $f$ is continuous on $[-3,3]$, it will achieve its maximum and minimum in $[-3,3]$. It can happen at the endpoints or at its critical values, which are $x=\tfrac{1}{\sqrt{2}}$ and $x=-\tfrac{1}{\sqrt{2}}$. So, $f$ attains its maximum or minimum on the set $\left\{-3,-\tfrac{1}{\sqrt{2}},\tfrac{1}{\sqrt{2}},3\right\}.$ Evaluate $f$ to see that

\[f(-3)\approx -0.00037,\quad f\left(-\tfrac{1}{\sqrt{2}}\right)\approx -0.42888,\quad f\left(\tfrac{1}{\sqrt{2}}\right)\approx 0.42888\quad\text{and}\quad f(6)\approx 0.00037.\]

So $f$ attains its minimum at $-\tfrac{1}{\sqrt{2}}$ and maximum at $\tfrac{1}{\sqrt{2}}$.

Part c

Because $f$ is continuous on $[-1,2]$, it will achieve its maximum and minimum in $[-1,2]$. It can happen at the endpoints or at its critical values (where $f^\prime=0$ or $f^\prime$ is undefined), which are $\{0,\tfrac{1}{3},1\}.$ So, $f$ attains its maximum or minimum on the set$ $\left\{-1,0,\tfrac{1}{3},1,2\right\}.$ Evaluate $f$ to see that

\[f(-1)=4,\quad f(0)=0,\quad f\left(-\tfrac{1}{3}\right)=\tfrac{4}{27},\quad f(1)=0,\quad \text{and}\quad f(2)=2.\]

So $f$ attains its minimum at both $0$ and $1$ and maximum at $-1$.

Mean Value Theorem

One of the most important results is called the mean value theorem.

Mean Value Theorem

Mean Value Theorem

For any real values $a$ and $b$ with $a$ less than $b$, and for any real valued function $f$, if

$f$ is differentiable on $(a,b)$,
$f$ is continuous on $[a,b]$,

then there is a $c$ in $(a,b)$ so that \[f^\prime(c) = \frac{f(b)-f(a)}{b-a}.\]

Heuristic: If a particle moves along a line in a time interval $[a,b]$, then there is a $c$ in $(a,b)$ at which the instantaneous velocity of the particle is equal to its average velocity on $[a,b]$.

Take $L$ to be the line that passes through $(a,f(a))$ and $(b, f(b))$.

The function $f-L$ is zero at $a$ and $b$, and $f-L$ is differentiable on $(a,b)$ and continuous on $[a,b]$.

Rolle’s theorem implies that there is a $c$ in $(a,b)$ such that

\[\begin{align*}(f-L)^\prime(c) &= f^\prime(c) - L^\prime(c) \\&= f^\prime(c) - \frac{f(b) - f(a)}{b-a}\\& = 0.\hspace{.41in}\end{align*}\]

The mean value theorem has many important consequences and the rest of this section is devoted to the study of these consequences.

Theorem on Differentiable Functions

For any function $f$, if that is differentiable on $(a,b)$ and continuous on $[a,b]$,

if $f$ is differentiable on $(a,b)$,
if $f$ is continuous on $[a,b]$,
if there is a natural number $M$ so that for any $z$ in $[a,b]$, \[|f^\prime(z)|\leq M,\]

then for any $x$ and $y$ in $[a,b]$, \[|f(x) - f(y)| \leq M|x-y|.\]

Here is why the theorem is true.

The mean value theorem guarantees that for any $x$ and $y$ in $[a,b]$, there is a $c$ between $x$ and $y$ so that \[f(x) - f(y) = f^\prime(c)(x-y),\]

and so \[|f(x) - f(y)| = |f^\prime(c)||x-y| \leq M|x-y|.\]

Example 4

Take $f(x)=\tfrac{1}{4}x^2+\frac{1}{2}$. This to a differentiable function on $(-1, 2)$ and continuous on $[-1,2]$.

Find a value $M$ so that $|f'(x)|\leq M$ for all $x$ in $[-1,2]$.
Given that $f(-1)=\tfrac{3}{4}$, find the smallest range of values that is guaranteed to contain $f([-1, 2])$ and sketch the smallest region that you can that is guaranteed to contain $f$. Then compare it to the graph of $f$.
Given that $f(-1)=\tfrac{3}{4}$ and $f(2)=\tfrac{3}{2}$, find the smallest range of values that is guaranteed to contain $f([-1, 2])$ and sketch the smallest region that you can that is guaranteed to contain $f$. Then compare it to the graph of $f$.

The derivative is $f'(x)=\frac{1}{2}x.$ This function is non-negative when $1\leq x<2$ and negative when $-1<x<1$. Therefore, the function $|f'(x)|$ has its highest absolute value at $f(2)=1.$ Therefore, $|f'(x)|\leq 1$ on $(-1,2).$
Use Theorem on Differentiable Functions to conclude that for $x\in(-1,2)$, $|f(x)-f(-1)|\leq 1 |x-(-1)|$ or $|f(x)-\tfrac{3}{4}|\leq |x+1|.$ Break up the absolute value of $|f(x)-\tfrac{3}{4}|$ to get \[f(x)-\tfrac{3}{4}\leq |x+1|\] when $f(x)\geq \tfrac{3}{4}$ and \[-\left(f(x)-\tfrac{3}{4}\right)\leq |x+1|\] when $f(x)<\tfrac{3}{4}$. Rewrite the two inequalities to get \[f(x)\leq |x+1|+\tfrac{3}{4}\] and \[f(x)\geq -|x+1|+\tfrac{3}{4}.\] This gives us a bound on the function.

Repeat the above process for $x\in(-1,2)$ with $|f(x)-f(2)|\leq 1 |x-2|$ to get that \[f(x)\leq |x-2|+\tfrac{3}{2}\] and \[f(x)\geq -|x-2|+\tfrac{3}{2}.\] Therefore \[f(x)\leq |x+1|+\tfrac{3}{4},\] \[f(x)\geq -|x+1|+\tfrac{3}{4},\] \[f(x)\leq |x-2|+\tfrac{3}{2},\] and \[f(x)\geq -|x-2|+\tfrac{3}{2}.\] The four bounds is some sort of 4 vertex polygon.

This theorem can also be used with paths as well.

Example 5

Take $c$ to be a path that is differentiable on the interval $[0,4]$ and for each $t$ in $[0,4]$, take $c(t)$ to be given by \[c(t) = (x(t), y(t)).\] For each of these choices of sampled values of $c$, given that for all $t$ in $I$, \[|x^\prime(t)|\leq 1\quad \text{and}\quad |y^\prime(t)|\leq 2,\] determine the smallest set $S(t)$ that is guaranteed to contain $c(t)$ for each $t$:

$c(0)= (3,-1)$;
$c(0)= (3,-1)$ and $c(4) = (5, 3)$.

Simulate $S(t)$ over the given time interval $I$.

Part a

For each $t$ in $[0,4]$, take $X(t)$ and $Y(t)$ to be, respectively, the set of possible positions for $x(t)$ and $y(t)$. Use the bounded derivative estimate with data $c(0)=(3,-1)=(x(0),y(0))$ on each component to get

\[-t+3\leq x(t)\leq t+3\quad\text{and}\quad -2t-1\leq y(t)\leq 2t-1\]

which implies that \[X(t)=[-t+3,t+3]\quad\text{and}\quad Y(t)=[-2t-1,2t-1].\]

The possible positions for $c(t)$ for each $t$ in $[0,4]$ is $X(t)\times Y(t)$. Here are sketches for the possible positions of each component.

Part b

\[-t+3\leq x(t)\leq t+3\quad\text{and}\quad t+1\leq x(t)\leq -t+9\] and \[-2t-1\leq y(t)\leq 2t-1\quad \text{and}\quad 2t-5\leq y(t)\leq -2t+11.\]

Here are some sketches of the region.

Use the sketches to see that

\[X(t)=\begin{cases}[-t+3,t+3]&\text{if }0\leq t<1\\ [t+1,t+3]&\text{if } 1\leq t<3\\ [t+1,-t+9]&\text{if }3\leq t\leq 4\end{cases}\quad\text{and}\quad Y(t)=\begin{cases}[-2t-1,2t-1]&\text{if }0\leq t<1\\ [2t-5,2t-1]&\text{if } 1\leq t<3\\ [2t-5,-2t+11]&\text{if }3\leq t\leq 4.\end{cases}\]

Having bounds on the derivative actually means that we can find a modulus of continuity.

Example 6

Take $f$ to be a differentiable function on an interval $I$ so that for any $x$ in $I$, \[f^\prime(x)| \leq M.\] For each of these choices of $M$, identify a modulus of continuity for $f$:

$M = 6$;
$M = \frac{1}{5}$.

Part a

Use the bounded derivative estimate to see that for any $x$ and $y$ in $I$, \[|f(x)-f(y)|\leq 6|x-y|\] so a modulus of continuity $\omega$ for $f$ is given by \[\omega(h)=6|h|.\]
Use the bounded derivative estimate to see that for any $x$ and $y$ in $I$, \[|f(x)-f(y)|\leq \tfrac{1}{5}|x-y|\] so a modulus of continuity $\omega$ for $f$ is given by \[\omega(h)=\tfrac{1}{5}|h|.\]

Antiderivatives and Fundamental Theorem of Calculus

Zero Derivative Theorem

For any interval $I$ and for any real valued function $f$ on $I$, if $f$ is differentiable on $I$, and if for every $x$ in $I$, \[f^\prime(x) = 0,\] then $f$ is a constant function.

This result is one of the most important consequences of the mean value theorem.

It may be thought of as a rigidity result that comes from the local linear approximation of a function: A function with the above condition on its derivative is entirely determined by its value at a single point.

Here is why this is true.

For any $x$ and $y$ in $I$, the mean value theorem implies that there is a $c$ between $x$ and $y$ so that \[f(x) - f(y) = f^\prime(c)(x-y) = 0\quad \text{and so}\quad f(x) = f(y).\]

Equal Derivative Theorem

For any interval $I$, and for any real valued functions $f$ and $g$ on $I$ that are differentiable on $I$, if for every $x$ in $I$, \[f^\prime(x) = g^\prime(x),\] then $f-g$ is a constant function.

Here is why this is true.

Linearity of the derivative implies that for any $x$ in $I$, \[(f-g)^\prime(x) = f^\prime(x) - g^\prime(x) = 0.\]

Since $(f-g)^\prime$ is equal to $0$ on $I$, the function $f-g$ is a constant function.

Example 7

Take $f$ and $g$ to be differentiable functions on the interval $[0,3]$ so that $f(1)=1$, $g(1)=2$, and that for any $x$ in $(0,3),$ \[f'(x)-g'(x)=0.\] Determine $g(3)-f(3).$

Because \[f'(x)-g'(x)=0\] for all $x$ in $(0,3)$ the function $h=f-g$ is constant.

That means for all $x$ in $[0,3]$, \[f(x)-g(x)=c\] for some constant $c$.

Because $f(1)=1$ and $g(1)=2$, \[\begin{align*} f(1)-g(1)&=c\\ 1-2&=c\\ -1&=c. \end{align*}\] Therefore, \[f(x)-g(x)=-1\] for all $x$ in $[0,3]$ so \[g(3)-f(3)=1.\]

Antiderivative

For any interval $I$ and any real valued function $f$ that is defined on $I$, if there is a function $F$ that is also defined on $I$, differentiable on $I$, and \[F^\prime = f,\] then $f$ has an antiderivative on $I$ and $F$ is an antiderivative of $f$ on $I$.

If $F$ and $G$ are both antiderivatives of $f$, then \[(F-G)^\prime = F^\prime - G^\prime = f-f = 0\] and so there is a constant function $C$ so that \[F = G + C.\]

Indefinite Integral

Define by the symbol \[\int f(x)\,{\rm d}x\] the indefinite integral of $f$, the set of functions that are antiderivatives of $f$ on $I$.

For any given antiderivative $F$ of $f$, write \[\int f(x)\,{\rm d}x = F(x)+C\] to indicate that any two functions that are antiderivatives of $f$ may differ only by a constant on the interval $I$.

The notation for the indefinite integral of a function, like the Leibniz notation for differentiation, allows us to work with formulas for function and not have to explicitly name functions.

Antidifferentiation is the operation that takes a function to its indefinite integral, although we sometimes abuse the term and talk about antidifferentiation as taking a function to an antiderivative of the function.

The various formulas for differentiation imply certain helpful formulas for antidifferentiation that we will explore below.

Such formulas can be helpful for determining antiderivatives, but the most fundamental way of checking that $F$ is an antiderivative for $f$ is to differentiate $F$.

Linearity of Antidifferentiation

Note that the linearity of differentiation also implies the linearity of antidifferentiation, that is: \[\int (af(x) +bg(x))\,{\rm d}x = a\int f(x)\,{\rm d}x + b\int g(x))\,{\rm d}x.\]

Example 8

For each function $f$ that is given below, determine the antiderivative of $f$:

$f(x) = 4x^2 + x + 5$;
$f(x) = \frac{5}{x} - 2\sqrt{x}$;
$f(x) = 2x + \sin(x)$;
$f(x) = \frac{7}{1+x^2}$
$f(x) = \frac{3}{\sqrt{1-x^2}} - 5$

First, if $n$ and $c$ are real numbers and $n+1\not= 0$, then the following are true:

$(x^{n+1})'=(n+1)x^{n}$,
$(cx)'=c$,
$(\cos(x))'=-\sin(x)$,
$(\arctan(x))'=\frac{1}{x^2+1}$,
$(\ln(x))'=\frac{1}{x},$ and
$(\arcsin(x))^\prime=\frac{1}{\sqrt{1-x^2}}.$

This means

$x^{n+1}+C=\int (n+1)x^{n}\,\mathrm{d}x$,
$cx+C=\int c\,\mathrm{d}x$,
$\cos(x)+C=\int (-\sin(x))\, \mathrm{d}x$,
$\arctan(x)+C=\int \frac{1}{x^2+1}\,\mathrm{d}x$
$\ln|x|+C=\int \frac{1}{x}\,\mathrm{d}x$ and
$\arcsin(x)+C=\int\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x.$

Rewrite to get the following antiderivative formulas

$\int x^{n}\,\mathrm{d}x=\frac{x^{n+1}}{n+1}+C$ for $n\not= -1$
$\int c\,\mathrm{d}x=cx+C$
$\int \sin(x)\, \mathrm{d}x=-\cos(x)+C$
$\int \frac{1}{x^2+1}\,\mathrm{d}x=\arctan(x)+C$
$\int \frac{1}{x}\,\mathrm{d}x=\ln|x|+C$ (so that $\ln$ is defined)
$\int\frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x=\arcsin(x)+C.$

Use linearity to get the following

\[\begin{align*}\int(4x^2 + x + 5)\,dx=&4\int x^2\,\mathrm{d}x+\int x\,\mathrm{d}x+\int 5\,\mathrm{d}x\\ &=4\left[\frac{x^{2+1}}{2+1}\right]+\left[\frac{x^{1+1}}{1+1}\right]+5x+C\\ &=\tfrac{4}{3}x^3+\tfrac{1}{2}x^2+5x+C\\\end{align*}\]
\[\begin{align*}\int\left(\frac{5}{x} - 2\sqrt{x}\right)\,dx=&5\int \frac{1}{x}\,\mathrm{d}x-2\int x^{\tfrac{1}{2}}\,\mathrm{d}x\\ &=5\ln(x)-2\left[\frac{x^{\frac{1}{2}+1}}{\frac{1}{2}+1}\right]+C\\ &=5\ln(x)-\tfrac{2}{\tfrac{3}{2}}x^{3/2}+C\\ &=5\ln(x)-\tfrac{4}{3}x^{3/2}+C\\\end{align*}\]
\[\begin{align*}\int\left(2x + \sin(x)\right)\,dx=&2\int x\,\mathrm{d}x+\int \sin(x)\,\mathrm{d}x\\ &=2\left[\frac{x^{1+1}}{1+1}\right]+\left[-\cos(x)\right]+C\\ &=x^2-\cos(x)+C\end{align*}\]
\[\begin{align*}\int\left(\frac{7}{1+x^2}\right)\,dx=&7\int \frac{1}{x^2+1}\,\mathrm{d}x\\ &=7\arctan(x)+C\end{align*}\]
\[\begin{align*} \int \left(\frac{3}{\sqrt{1-x^2}} - 5\right)\,\mathrm{d}x &=3\int \frac{1}{\sqrt{1-x^2}}\,\mathrm{d}x-5\int 1\,\mathrm{d}x\\ &=3\arcsin(x)-5x+C, \end{align*}\] Check these are valid by taking the derivative of the answer and confirming the original function is obtained.

We have the following theorem that allows us to calculate indefinite integrals:

Fundamental Theorem of Calculus I (FTC 1)

Fundamental Theorem of Calculus I (FTC 1) If $F$ is differentiable on the interval $[a,b]$ and if for each $x$ in $[a,b]$, \[F'(x)=f(x)\] and if $f$ is integrable on $[a,b]$, then \[\int_a^bf(x)\,{\rm{d}}x=F(b)-F(a).\] Another notation for this is given by \[F(x)\biggr|^b_a=F(b)-F(a).\]

Example 9

Calculate the following:

$\displaystyle\int_{0}^1x^2\,\mathrm{d}x$
$\displaystyle\int_{-2}^3 (4x^2 + x + 5)\,\mathrm{d}x$

First note that $\int x^2\,\mathrm{d}x=\tfrac{1}{3}x^3+C$, so $F(x)=\tfrac{1}{3}x^3+C$ is an antiderivative of $f(x)=x^2$. By the Fundamental Theorem of Calculus I, \[\int_0^1x^2\,\mathrm{d}x=(\tfrac{1}{3}x^3+C)\biggr|^1_0=\tfrac{1}{3}(1)^3+C-\tfrac{1}{3}(0)^3-C=\tfrac{1}{3}\]
First note that $\int (4x^2 + x + 5)\,\mathrm{d}x=\tfrac{4}{3}x^3+\tfrac{1}{2}x^2+5x+C$, so $F(x)=\tfrac{4}{3}x^3+\tfrac{1}{2}x^2+5x+C$ is an antiderivative of $f(x)=4x^2 + x + 5$. By the Fundamental Theorem of Calculus I, \[\int_{-2}^3 (4x^2 + x + 5)\,\mathrm{d}x=(F(x))\biggr|^3_{-2}=F(3)-F(-2)=\tfrac{111}{2}+\tfrac{56}{3}\]

As we seen above, $C$ won’t matter because of the cancellation. So for definite integrals, we usually let $C=0$.

Next result allows us to take the derivative of a function defined using a definite integral.

Fundamental Theorem of Calculus II (FTC 1I)

Fundamental Theorem of Calculus II (FTC 1I) If $f$ is continuous on $[a,b]$, then $F(x)=\int_a^xf(t)\,\mathrm{d}t$ is differentiable on $(a,b)$ and \[F'(x)=\left(\int_a^xf(t)\,\mathrm{d}t\right)'=f(x)\]

Example 10

The derivative of $F(x)=\int_3^x \sin(x)\,\mathrm{d}x$ is $F'(x)=\sin(x)$.
The derivative of $F(x)=\int_3^{x^2} \ln(x)e^x\,\mathrm{d}x$ is $F'(x)=\ln(x^2)e^{x^2}\cdot 2x$. (chain rule)

Reverse Chain Rule or U-Substitution

One of the ways we can determine antiderivatives is by recognizing certain forms. For example, something that resembles the result of taking the derivative of a composite function.

Reverse Chain Rule

For any differentiable functions $F$ and $G$, if for any $x$ in an interval $I$ and any $z$ in $g(I)$, \[F^\prime(z) = f(z)\quad\text{and}\quad G^\prime(x) = g(x),\] we have

\[\int f(G(x))g(x)\,{\rm d}x = F(G(x)) + C.\]

This formula is the reverse chain rule. Some call it $u$-substitution because if we let $u(x)=G(x)$, then the theorem can be rewritten like this

\[\int f(u(x))u'(x)\,{\rm d}x = F(u(x)) + C.\]

This is why this is true.

The chain rule for differentiation implies that \[(F\circ G)^\prime(x) = f(G(x))g(x).\]

The mean value theorem implies that \[\int (F\circ G)^\prime(x) \,{\rm d}x = F(G(x)) + C\quad\text{and so}\quad \int f(G(x))g(x)\,{\rm d}x = F(G(x)) + C.\]

Example 11

Determine the following antiderivatives:

$\displaystyle \int 2x\sin(x^2)\,{\rm d}x$;
$\displaystyle \int^3_2 (4x^3 +3)(x^4 + 3x -5)^{\frac{2}{3}}\,{\rm d}x$;
$\displaystyle \int \frac{2x}{x^2 +3}\,{\rm d}x$;
$\displaystyle \int (3x^2+1)\cos(x^3+x+2)\,{\rm d}x$.

Notice that $(\sin(x))'=\cos(x)$ so $\sin(x)+C=\int \cos(x)\,\mathrm{d}x.$

Let $f(x)=\sin(x)$ and $u(x)=x^2$. Then $u'(x)=2x$. Notice that $2x\sin(x^2)=2xf(x^2)=2xf(u)=f(u)2x=f(u)u'.$ So by the reverse chain rule $\displaystyle \int 2x\sin(x^2)\,{\rm d}x=F(u)+C$ where $F$ is an antiderivative of $f(x)=\sin(x)$ and $C$ is a constant. So in this case $F(x)=-\cos(x)+C$. Thus $\displaystyle \int 2x\sin(x^2)\,{\rm d}x=F(u)+C=-\cos(u)+C=-\cos(x^2)+C.$
Let $f(x)=x^{2/3}$ and $u(x)=x^4+3x-5$. Then $u'(x)=4x^3+3$. Notice that $(4x^3 +3)(x^4 + 3x -5)^{\frac{2}{3}}=(4x^3+3)u^{2/3}=f(u)(4x^3+3)=f(u)u'.$ So by the reverse chain rule $\displaystyle \int (4x^3 +3)(x^4 + 3x -5)^{\frac{2}{3}}\,{\rm d}x=F(u)+C$, where $F$ is an antiderivative of $f(x)=x^{2/3}$ and $C$ is a constant. So in this case $F(x)=\frac{3}{5}x^{5/3}+C$. Thus $\displaystyle \int (4x^3 +3)(x^4 + 3x -5)^{\frac{2}{3}}\,{\rm d}x=F(u)+C=\tfrac{3}{5}(u)^{5/3}+C=\tfrac{3}{5}(x^4+3x-5)^{5/3}+C$. To compute \[\displaystyle \int^3_2 (4x^3 +3)(x^4 + 3x -5)^{\frac{2}{3}}\,{\rm d}x\] use the antiderivative $G(x)=\tfrac{3}{5}(x^4+3x-5)^{5/3}+C$ and the fundamental theorem of calculus I to get \[\displaystyle \int^3_2 (4x^3 +3)(x^4 + 3x -5)^{\frac{2}{3}}\,{\rm d}x=G(3)-G(2)=\tfrac{3}{5}(85)^{5/3}-\tfrac{3}{5}(17)^{5/3}\]
Let $u(x)=x^2+3$. Notice that $u'(x)=2x$. So $\frac{2x}{x^2+3}=\frac{u'}{u}=(u)^{-1}u'$. Let $f(x)=\frac{1}{x}$. Then we see that $\frac{2x}{x^2+3}=f(u)u'.$ So by the reverse chain rule $\displaystyle \int \frac{2x}{x^2 +3}\,{\rm d}x=F(u)+C$, where $F$ is an antiderivative of $f(x)=\frac{1}{x}$. In this case, $F(x)=\ln|x|,$ The final answer is $\displaystyle \int \frac{2x}{x^2 +3}\,{\rm d}x=\ln|x^2+3|+C$
Let $f(x)=\cos(x)$ and $u(x)=x^3+x+2$. Then $u'(x)=3x^2+1$. Notice that $(3x^2+1)\cos(x^3+x+2)=(3x^2+1)f(x^3+x+2)=(3x^2+1)f(u)=f(u)(3x^2+1)=f(u)u'.$ So by the reverse chain rule $\displaystyle \int (3x^2+1)\cos(x^3+x+2)\,{\rm d}x=F(u)+C$ where $F$ is an antiderivative of $f(x)=\cos(x)$ and $C$ is a constant. So in this case $F(x)=\sin(x)+C$. Thus $\displaystyle\int (3x^2+1)\cos(x^3+x+2)\,{\rm d}x=F(u)+C=\sin(u)+C=\sin(x^3+x+2)+C.$

Reverse Product Rule

The product rule states that

\[(f(x)g(x))^\prime = f^\prime(x)g(x)+f(x)g^\prime(x).\]

Integrating both sides gives us

\[f(x)g(x)+C=\int f^\prime(x)g(x)\,\mathrm{d}x+\int f(x)g^\prime(x)\,\mathrm{d}x\]

which is equivalent to

\[\int f^\prime(x)g(x)\,\mathrm{d}x=f(x)g(x)-\int f(x)g^\prime(x)\,\mathrm{d}x+C.\]

This is known as the reverse product rule.

Use this to determine the following indefinte integrals.

Example 12

For each of these choices of function $f$, determine $\int f(x) \,{\rm d}x$ and identify an antiderivative of $f$:

$f(x) = x\sin(x)$;
$f(x) = x\ln(|x|)$;
$f(x) = x{\rm e}^x$.

Part a

Notice that \[(-x\cos(x))^\prime=-\cos(x)+x\sin(x)\] and so

\[\int(-x\cos(x))^\prime\,\mathrm{d}x=-\int\cos(x)\,\mathrm{d}x+\int x\sin(x)\,\mathrm{d}x.\]

Rearrange this equality to see that

\[\begin{align*} \int x\sin(x)\,\mathrm{d}x &=\int(-x\cos(x))^\prime\,\mathrm{d}x+\int \cos(x)\,\mathrm{d}x\\ &=-x\cos(x)-\sin(x)+C. \end{align*}\]

An antiderivative of $f$ is given by \[F(x)=-\sin(x)-x\cos(x)+57.\]

Part b

Notice that \[(x^2\ln|x|)^\prime=2x\ln(|x|)+x\] and so

\[\int(x^2\ln|x|)^\prime \,\mathrm{d}x=2\int x\ln(|x|)\,\mathrm{d}x+\int x\,\mathrm{d}x.\]

Rearrange this equality to see that

\[\begin{align*} \int x\ln(|x|)\,\mathrm{d}x &=\tfrac{1}{2}\int (x^2\ln|x|)^\prime \,\mathrm{d}x-\tfrac{1}{2}\int x\,\mathrm{d}x \\ &=\tfrac{1}{2}x^2\ln(|x|)-\tfrac{1}{4}x^2+C. \end{align*}\]

An antiderivative of $f$ is given by \[F(x)=\tfrac{1}{2}x^2\ln(|x|)-\tfrac{1}{4}x^2+5.\]

Part b

Notice that \[(x^2\ln|x|)^\prime=2x\ln(|x|)+x\] and so

\[\int(x^2\ln|x|)^\prime \,\mathrm{d}x=2\int x\ln(|x|)\,\mathrm{d}x+\int x\,\mathrm{d}x.\]

Rearrange this equality to see that

\[\begin{align*} \int x\ln(|x|)\,\mathrm{d}x &=\tfrac{1}{2}\int (x^2\ln|x|)^\prime \,\mathrm{d}x-\tfrac{1}{2}\int x\,\mathrm{d}x \\ &=\tfrac{1}{2}x^2\ln(|x|)-\tfrac{1}{4}x^2+C. \end{align*}\]

An antiderivative of $f$ is given by \[F(x)=\tfrac{1}{2}x^2\ln(|x|)-\tfrac{1}{4}x^2+5.\]

Part c

Notice that \[\left(x\mathrm{e}^x\right)^\prime=\mathrm{e}^x+x\mathrm{e}^x\] and so

\[\int\left(x\mathrm{e}^x\right)^\prime \,\mathrm{d}x=\int \mathrm{e}^x\,\mathrm{d}x+\int x\mathrm{e}^x\,\mathrm{d}x.\]

Rearrange this equality to see that

\[\begin{align*} \int x\mathrm{e}^x\,\mathrm{d}x &=\int \left(x\mathrm{e}^x\right)^\prime \,\mathrm{d}x-\int \mathrm{e}^x\,\mathrm{d}x \\ &=x\mathrm{e}^x-\mathrm{e}^x+C. \end{align*}\]

An antiderivative of $f$ is given by \[F(x)=x\mathrm{e}^x-\mathrm{e}^x.\]

When determining an antiderivative, it is helpful to pay attention to the form of the solution and attempt to find candidates that have a certain form.

Differentiating the form of the antiderivative can help us to determine the actual antiderivative.

The function $f(x)$ is said to be the integrand in an indefinite integral $\int f(x)\,{\rm d}x$—To determine indefinite integral, study the form and structure of its integrand.

It may help to use linearity of antidifferentiation to simplify an indefinite integral.

Example 13

Evaluate each of these indefinite integrals:

$\displaystyle \int \tfrac{5x-3}{x^2 +1}\,{\rm d}x$;
$\displaystyle \int (2x+1)\sin(x)\,{\rm d}x$;
$\displaystyle \int \tfrac{1}{x\ln(|x|)}\,{\rm d}x$;
$\displaystyle \int 7x\sec^2(x^2+1)\,{\rm d}x$.

Part a

Use linearity of antidifferentiation and the reverse chain rule to see that

\[\begin{align*} \int \frac{5x-3}{x^2+1}\,\mathrm{d}x &=5\int \frac{x}{x^2+1}\,\mathrm{d}x-3\int \frac{1}{x^2+1}\,\mathrm{d}x\\ &=\tfrac{5}{2}\int (x^2+1)^\prime \ln^\prime |x^2+1|\,\mathrm{d}x-3\int \frac{1}{x^2+1}\,\mathrm{d}x\\ &=\tfrac{5}{2}\int (\ln|x^2+1|)^\prime\,\mathrm{d}x-3\int \frac{1}{x^2+1}\,\mathrm{d}x\\ &=\tfrac{5}{2}\ln|x^2+1|-3\arctan(x)+C. \end{align*}\]

Part b

Use linearity of antidifferentiation and the equality \[(-x\cos(x))^\prime=-\cos(x)+x\sin(x)\] to see that

\[\begin{align*} \int (2x+1)\sin(x)\,\mathrm{d}x &=2\int x\sin(x)\,\mathrm{d}x+\int \sin(x)\,\mathrm{d}x\\ &=2\int \left((-x\cos(x))^\prime+\cos(x)\right)\,\mathrm{d}x+\int \sin(x)\,\mathrm{d}x\\ &=2\int (-x\cos(x))^\prime\,\mathrm{d}x+2\int \cos(x)\,\mathrm{d}x+\int \sin(x)\,\mathrm{d}x\\ &=-2x\cos(x)+2\sin(x)-\cos(x)+C. \end{align*}\]

Part c

Use the reverse chain rule to see that

\[\begin{align*} \int \frac{1}{x\ln(|x|)}\,\mathrm{d}x &=\int \frac{1}{x}\cdot\frac{1}{\ln(|x|)}\,\mathrm{d}x\\ &=\int (\ln(|x|))^\prime \cdot\ln^\prime|\ln(|x|)|\,\mathrm{d}x\\ &=\int \left(\ln|\ln(|x|)|\right)^\prime\,\mathrm{d}x\\ &=\ln(\ln(|x|))+C. \end{align*}\]

Part d

Use the reverse chain rule and $\left(\tan(x)\right)^\prime=\sec^2(x)$ to see that

\[\begin{align*} \int 7x\sec^2(x^2+1)\,\mathrm{d}x &=\tfrac{7}{2}\int 2x\sec^2(x^2+1)\,\mathrm{d}x\\ &=\tfrac{7}{2}\int (x^2+1)^\prime\tan^\prime(x^2+1)\,\mathrm{d}x\\ &=\tfrac{7}{2}\int \left(\tan(x^2+1)\right)^\prime \,\mathrm{d}x\\ &=\tfrac{7}{2}\tan(x^2+1)+C. \end{align*}\]

We can use antidifferentiation to recover motion if we have information on the velocity and data on where the particle was at.

Example 14

For each of these choices of vector valued function $v$, the function $v(t)$ is the velocity of a path $c$ at time $t$. For these choices of $t_0$ and $c(t_0)$, reconstruct the path $c$ and simulate the motion of the particle together with the particle’s velocity vector:

$v(t) = \langle 3t +1, -1\rangle$ and $c(1) = (6,3)$;
$v(t) = \langle -3\sin(3t), 6\cos(3t)\rangle$ and $c(0) = (3,0)$.

Part a

Take $x$ and $y$ to be functions so that

\[x^\prime (t)=3t+1\quad\text{and}\quad y^\prime(t)=-1.\]

An antiderivative for $x$ and $y$ are given by

\[x(t)=\tfrac{3}{2}t^2+t+x_0\quad\text{and}\quad y(t)=-t+y_0\]

and so the path $c$ is of the form

\[c(t)=\left(\tfrac{3}{2}t^2+t+x_0,-t+y_0\right).\]

Use the data point $c(1)=(6,3)$ to see that

\[\begin{align*} c(1)&=(6,3)\\ \left(\tfrac{3}{2}(1)^2+(1)+x_0,-(1)+y_0\right)&=(6,3)\\ \left(\tfrac{5}{2}+x_0,-1+y_0\right)&=(6,3), \end{align*}\]

which implies that \[x_0=\tfrac{7}{2}\quad\text{and}\quad y_0=4.\]

Thus the reconstructed path is

\[c(t)=\left(\tfrac{3}{2}t^2+t+\tfrac{7}{2},-t+4\right).\]

Part b

Take $x$ and $y$ to be functions so that

\[x^\prime (t)=-3\sin(3t)\quad\text{and}\quad y^\prime(t)=6\cos(3t).\]

An antiderivative for $x$ and $y$ are given by

\[x(t)=\cos(3t)+x_0\quad\text{and}\quad y(t)=2\sin(3t)+y_0\]

and so the path $c$ is of the form

\[c(t)=\left(\cos(3t)+x_0,2\sin(3t)+y_0\right).\]

Use the data point $c(0)=(3,0)$ to see that

\[\begin{align*} c(1)&=(3,0)\\ \left(\cos(3(0))+x_0,2\sin(3(0))+y_0\right)&=(3,0)\\ \left(1+x_0,y_0\right)&=(3,0), \end{align*}\]

which implies that \[x_0=2\quad\text{and}\quad y_0=0.\]

Thus the reconstructed path is

\[c(t)=\left(\cos(3t)+2,2\sin(3t)\right).\]

L’Hoptial’s Rule

For any interval $I$ that contains $x_0$, and any functions $f$ and $g$ defined on $I\setminus\{x_0\}$, the following limits are said to have an indeterminate form:

($\tfrac{0}{0}$ or $\pm\tfrac{\infty}{\infty}$)

$\displaystyle \lim_{x\to x_0} \tfrac{f(x)}{g(x)}$ and either $\displaystyle \lim_{x\to x_0} f(x) = \lim_{x\to x_0} g(x) = 0$ or $\displaystyle \lim_{x\to x_0} f(x) = \pm\infty$ and $\displaystyle\lim_{x\to x_0} g(x) = \pm\infty$

($\infty - \infty$)

$\displaystyle \lim_{x\to x_0} (f(x) - g(x))$ and either $\displaystyle \lim_{x\to x_0}f(x)=\lim_{x\to\infty}g(x)=\infty$ or $\displaystyle \lim_{x\to x_0} f(x) = \lim_{x\to x_0} g(x) = -\infty$.

($\infty^0$, $1^{\pm\infty}$,$0^0$)

$\displaystyle \lim_{x\to x_0} (f(x))^{g(x)}$ and either $\displaystyle \lim_{x\to x_0} f(x) = \infty$ and $\displaystyle \lim_{x\to x_0} g(x) = 0$, or $\displaystyle \lim_{x\to x_0} f(x) = 1$ and $\displaystyle \lim_{x\to x_0} g(x) = \pm\infty$ or $\displaystyle \lim_{x\to x_0} f(x) = 0$, $f$ is non-negative, and $\displaystyle \lim_{x\to x_0} g(x) = 0$.

Example 15

Identify which of these limits have indeterminate forms:

$\displaystyle \lim_{x \to 0} \tfrac{\sin(x^2)}{x}$;
$\displaystyle \lim_{x \to \frac{1}{2}} \tfrac{\cos(\pi x)}{(x-\frac{1}{2})}$;
$\displaystyle \lim_{x\to 1} \tfrac{x^2 +1}{{\rm e}^x}$;
$\displaystyle \lim_{x\to \infty} \left(\sqrt{x+1} - \sqrt{x}\right)$;
$\displaystyle \lim_{x\to 3} \tfrac{x}{(x-3)^2}$.
$\displaystyle \lim_{x\to 4} \left(\sqrt{x+1} - \sqrt{x}\right)$;
$\displaystyle \lim_{x\to \frac{1}{2}^+} \tfrac{\ln(x-\frac{1}{2})}{\tan(\pi x)}$;
$\displaystyle \lim_{x\to 0} x\ln(x^2)$;
$\displaystyle \lim_{x\to 0^+} x^x$;
$\displaystyle \lim_{x\to \infty} \tfrac{x^2}{{\rm e}^x}$;
$\displaystyle \lim_{x\to \infty} \tfrac{{\rm e}^{-x}}{\arctan(x)}$.

0/0
0/0
Not indeterminant because limit of numerator and denominator are both non-zero.
$\infty -\infty$
Not indeterminant form because the numerator has a finite non-zero limit
Not indeterminant because both limits are finite
$\infty/\infty$
Rewrite as $\frac{x}{\frac{1}{\ln(x^2)}}$ then we have $0/0$
$0^0$
$\infty/\infty$
Not indeterminant form since denominator has a non-zero limit.

L’Hoptial’s Rule

L’Hoptial’s Rule

For any interval $(a,b)$, any $x_0$ in $(a,b)$ and any functions $f$ and $g$ that are continuous on $[a,x_0)\cup(x_0,b]$, differentiable on $(a,x_0)\cup(x_0,b)$ and either \[\lim_{x\to x_0}f(x)=\lim_{x\to x_0}g(x)=0,\] or \[\lim_{x\to x_0}f(x)=\pm\infty \text{ or }\lim_{x\to x_0}g(x)=\pm\infty,\] then if \[\lim_{x\to x_0}\frac{f'(x)}{g'(x)}\] exists, then so does \[\lim_{x\to x_0}\frac{f(x)}{g(x)}\] and both limits are equal.

Now we practice computing some of these.

Example 16

Verify that the following are indeterminate forms and use L’Hopital’s rule to determine the given limits:

$\displaystyle \lim_{x\to 3}\frac{x^2-x-6}{x-3}$;
$\displaystyle \lim_{x\to -2}\frac{2x^3+4x^2+x+2}{x+2}$;
$\displaystyle \lim_{x\to \infty}\left(\sqrt{x^2 +3x + 1} -x\right)$;
$\displaystyle\lim_{x\to 0^+} x\ln(x)$;
$\displaystyle \lim_{x\to 0^+} x^{x}$;
$\displaystyle \lim_{x\to \infty} \tfrac{x}{{\rm e}^x}$;
$\displaystyle \lim_{x\to \infty} \tfrac{x^2}{{\rm e}^x}$;

$\displaystyle\lim_{x\to 3}(x^2-x-6)=0$ and $\displaystyle\lim_{x\to 3}(x-3)=0$ so indeterminant form $0/0$. Use L’H rule to get \[\begin{align*}\displaystyle \lim_{x\to 3}\frac{x^2-x-6}{x-3}&=\lim_{x\to 3}\frac{(x^2-x-6)'}{(x-3)'}\\ &=\lim_{x\to 3}\frac{2x-1}{1-0}\\ &=\lim_{x\to 3}(2x-1)\\ &=6-1\\ &=5\end{align*}\]
$\displaystyle\lim_{x\to -2}(2x^3+4x^2+x+2)=0$ and $\displaystyle\lim_{x\to -2}(x+2)=0$ so indeterminant form $0/0$. Use L’H rule to get \[\begin{align*}\displaystyle \lim_{x\to -2}\frac{2x^3+4x^2+x+2}{x+2}&=\lim_{x\to -2}\frac{(2x^3+4x^2+x+2)'}{(x+2)'}\\ &=\lim_{x\to -2}\frac{6x^2+8x+1}{1}\\ &=\lim_{x\to -2}(6x^2+8x+1)\\ &=24-16+1\\ &=9\end{align*}\]
$\displaystyle\lim_{x\to \infty}\sqrt{x^2 +3x + 1}=\infty$ and $\displaystyle\lim_{x\to \infty}x=\infty$ so indeterminant form of $\infty-\infty$. Need to rewrite as a quotient. Use conjugate trick: \[\begin{align*}\displaystyle \lim_{x\to \infty}\left(\sqrt{x^2 +3x + 1} -x\right)&=\displaystyle \lim_{x\to \infty}\left(\sqrt{x^2 +3x + 1} -x\right)\cdot \frac{\sqrt{x^2+3x+1}+x}{\sqrt{x^2+3x+1}+x}\\ &=\displaystyle \lim_{x\to \infty}\frac{x^2+3x+1-x^2}{\sqrt{x^2+3x+1}+x}\\ &=\displaystyle \lim_{x\to \infty}\frac{3x+1}{\sqrt{x^2+3x+1}+x}\\ &=\displaystyle \lim_{x\to \infty}\frac{(3x+1)'}{(\sqrt{x^2+3x+1}+x)'}&\text{ L'H Rule}\\ &=\displaystyle \lim_{x\to \infty}\frac{3}{\frac{2x+3}{2\sqrt{x^2+3x+1}}+1}\\ \end{align*}\] The limit of $\frac{2x+3}{2\sqrt{x^2+3x+1}}$ can computed by multiplying top and bottom by $\frac{1}{x}$ or $\frac{1}{\sqrt{x^2}}$. The limit will end up equaling $1$. Thus \[\begin{align*}\displaystyle \lim_{x\to \infty}\left(\sqrt{x^2 +3x + 1} -x\right)&=\displaystyle \lim_{x\to \infty}\left(\sqrt{x^2 +3x + 1} -x\right)\cdot \frac{\sqrt{x^2+3x+1}+x}{\sqrt{x^2+3x+1}+x}\\ &=\displaystyle \lim_{x\to \infty}\frac{x^2+3x+1-x^2}{\sqrt{x^2+3x+1}+x}\\ &=\displaystyle \lim_{x\to \infty}\frac{3x+1}{\sqrt{x^2+3x+1}+x}\\ &=\displaystyle \lim_{x\to \infty}\frac{(3x+1)'}{(\sqrt{x^2+3x+1}+x)'}&\text{ L'H Rule}\\ &=\displaystyle \lim_{x\to \infty}\frac{3}{\frac{2x+3}{2\sqrt{x^2+3x+1}}+1}\\ &=\frac{3}{1+1}\\ &=\frac{3}{2}. \end{align*}\]
Rewrite $x\ln(x)=\frac{\ln(x)}{\frac{1}{x}}.$ Then the limit is of the form $\infty/\infty$. Use L’H rule to get \[\begin{align*}\displaystyle \lim_{x\to 0^+} x\ln(x)&=\displaystyle \lim_{x\to 0^+}\frac{\ln(x)}{\frac{1}{x}}\\ &=\displaystyle \lim_{x\to 0^+}\frac{(\ln(x))'}{(\frac{1}{x})'}&\text{ L'H Rule}\\ &=\displaystyle \lim_{x\to 0^+}\frac{\frac{1}{x}}{-\frac{1}{x^2}}\\ &=\displaystyle \lim_{x\to 0^+}\frac{1}{x}\cdot(-2x^2)\\ &=\displaystyle \lim_{x\to 0^+}(-2x)\\ &=0. \end{align*}\]
This is of the form $0^0$. Rewrite the expression as a quotient: $x^x=e^{\ln(x^x)}=e^{x\ln(x)}$. Use the continuity of exponential function, to conclude that to calculate the limit of $x^x,$ calculate instead $e^{\displaystyle\lim_{x\to 0^+}x\ln(x)}$. Thus $\displaystyle \lim_{x\to 0^+}x^x=e^{\displaystyle\lim_{x\to 0^+}x\ln(x)}=e^0=1$.
The form is $\infty/\infty,$ so use L’H rule \[\begin{align*}\displaystyle \lim_{x\to \infty} \frac{x}{e^x}&=\displaystyle \lim_{x\to \infty}\frac{(x)'}{(e^x)'}\\ &=\displaystyle \lim_{x\to \infty}\frac{1}{e^x}\\ &=0. \end{align*}\]
The form is $\infty/\infty,$ so use L’H rule \[\begin{align*}\displaystyle \lim_{x\to \infty} \frac{x^2}{e^x}&=\displaystyle \lim_{x\to \infty}\frac{(x^2)'}{(e^x)'}&\text{L'H Rule}\\ &=\displaystyle \lim_{x\to \infty}\frac{2x}{e^x}\\ &=\displaystyle \lim_{x\to \infty}\frac{(2x)'}{(e^x)'}&\text{L'H Rule}\\ &=\displaystyle \lim_{x\to \infty}\frac{2}{e^x}\\ &=0. \end{align*}\]

L’Hopital’s Rule does not necessarily simplify the determination of a limit.

Example 17

Observe what happens if you attempt to use L’Hopital’s rule to calculate \[\lim_{x\to 0} \frac{\sqrt{x^2+1}}{x}\]

L’H rule does not apply because the numerator has a finite non-zero limit. Improper use of L’H rule results in \[\lim_{x\to 0} \frac{\sqrt{x^2+1}}{x}=\lim_{x\to 0}\frac{\frac{2x}{2\sqrt{x^2+1}}}{1}=\lim_{x\to 0}\frac{2x}{2\sqrt{x^2+1}}=\frac{0}{1}=0.\] This is false, see the graph.

Example 18

Determine what is incorrect about these statements:

L’Hopital’s rule implies that \[\displaystyle \lim_{x\to \infty} \frac{1+3x}{2x+\sin(x)} = \lim_{x\to \infty} \frac{3}{2+\cos(x)}.\] Since the second limit does not exist, neither does the first.
L’Hopital’s rule implies that \[\displaystyle \lim_{x\to \infty} \dfrac{{\rm e}^x - 1}{x^2} = \lim_{x\to \infty} \dfrac{{\rm e}^x}{2x}.\] Since the second limit does not exist, neither does the first.

A conclusion can only be made if the new limit exist. If the limit does not exist, nothing can be said about the original limit.
The newer limit does exist. Need to take L’H rule again.

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