Chapter 5.5 Practice

Questions

  1. Use the sequential definition of a limit to determine which of the limits below exist and which do not exist.

    1. \(\displaystyle\lim_{x\to -3}(x^3+2x+11)\)
    2. \(\displaystyle\lim_{x\to 4}\frac{x^2-16}{x-4}\)
    3. \(\displaystyle\lim_{x\to 0}\cos\left(\frac{1}{x}\right)\)

  2. Determine the following limits:

    1. \(\displaystyle\lim_{x\to 9}4\sqrt{x}\)
    2. \(\displaystyle\lim_{x\to 4}\left(11+\frac{1}{x}\right)\)
    3. \(\displaystyle\lim_{x\to 1}\frac{\cos(x-1)+2x^3}{(x+1)^2}\)
    4. \(\displaystyle\lim_{x\to 6}\sqrt{x^2+1}\)
    5. \(\displaystyle\lim_{x\to 6}\frac{x-6}{x^2+36}\)
    6. \(\displaystyle\lim_{x\to 6}\frac{x-6}{x^2-36}\)
    7. \(\displaystyle\lim_{x\to 0}\frac{\sin(4x)}{10x}\)
    8. \(\displaystyle\lim_{x\to 0}\frac{\tan(x)}{10x}\)

  3. Determine the following limit by using squeeze theorem

    1. \(\displaystyle\lim_{x\to 0}10x^2\sin\left(\frac{1}{x}\right)\)
    2. \(\displaystyle\lim_{x\to 3^-}(x-3)f(x),\) where \(f\) is a function defined on \((-\infty,3)\cup(3,\infty)\) and \(\frac{1}{x-3}\leq f(x)\leq \frac{2^{x-3}}{x-3}\) for all \(x\not=3.\)
    3. \(\displaystyle\lim_{x\to 0}5x^2e^{\sin\left(\frac{1}{x}\right)}\)

  4. Let \(g\) be a function defined on \([-6,3)\cup(3,8]\) whose graph is given below. Calculate the following:

    1. \(\displaystyle\lim_{x\to -4^-}g(x)\)
    2. \(\displaystyle\lim_{x\to -4^+}g(x)\)
    3. \(\displaystyle\lim_{x\to -4}g(x)\)
    4. \(g(-4)\)
    5. Is \(\displaystyle\lim_{x\to -4}g(x)=g(-4)?\)
    6. \(\displaystyle\lim_{x\to 1^-}g(x)\)
    7. \(\displaystyle\lim_{x\to 1^+}g(x)\)
    8. \(\displaystyle\lim_{x\to 1}g(x)\)
    9. \(\displaystyle g(1)\)
    10. Is \(\displaystyle\lim_{x\to 1}g(x)=g(1)?\)
    11. \(\displaystyle\lim_{x\to 6^-}g(x)\)
    12. \(\displaystyle\lim_{x\to 6^+}g(x)\)
    13. \(\displaystyle\lim_{x\to 6}g(x)\)
    14. \(\displaystyle g(6)\)
    15. Is \(\displaystyle \lim_{x\to 6}g(x)=g(6)?\)

  5. Take \(f\) to be the function that is given by \[f(x)=\begin{cases}\frac{x-2}{x^2-4}&\text{if }x<2\\\log_{16}(x)&\text{if }x\geq 2\end{cases}.\]

    1. Evaluate \(\displaystyle\lim_{x\to 2^{-}}f(x)\).
    2. Evaluate \(\displaystyle\lim_{x\to 2^{+}}f(x)\).
    3. Does \(\displaystyle\lim_{x\to 2}f(x)\) exist? If so, evaluate this limit. If not, explain why.

  6. Take \(b\) to be a real number and \(f\) to be the function that is given by \[f(x)=\begin{cases}\exp_3(x)&\text{if }x\leq 3\\bx^2+4&\text{if }x> 3\end{cases}.\] Determine a value for \(b\) so that \(\displaystyle\lim_{x\to 2}f\) exists.

  7. Evaluate the following limits. If they do not exist, state so.

    1. \(\displaystyle\lim_{x\to\infty}\frac{9x^2-20x+4}{15x^2+x+1}\)
    2. \(\displaystyle\lim_{x\to\infty}\frac{x^6+x^5+3x^2+4x+1}{3x^4+x^2+x+1}\)
    3. \(\displaystyle\lim_{x\to\infty}(3\cos(4x))\)
    4. \(\displaystyle\lim_{x\to\infty}\frac{3x}{x^2+3x+1}\)
    5. \(\displaystyle\lim_{x\to\infty}5^{x}+2\)
    6. \(\displaystyle\lim_{x\to-\infty}5^{x}+2\)
    7. \(\displaystyle\lim_{x\to\infty}\frac{1}{x}\sin(x)\)
    8. \(\displaystyle\lim_{x\to-\infty}\frac{1}{x}\sin(x)\)

  8. For each rational function \(f\), determine all vertical and horizontal asymptotes. Write your answers using the appropriate limit notation. It may be helpful to sketch \(f\).

    1. \(\displaystyle f(x)=\frac{(x-2)(x+4)}{x-3}\)
    2. \(\displaystyle f(x)=\frac{(x-2)}{x(x+4)}\)
    3. \(\displaystyle f(x)=\frac{6x-2}{3x+5}\)

  9. Take \(c\) to be the path that is given by \[c(t)=\left(3t-1,(t-2)\sin\left(\frac{1}{t-2}\right)\right).\] Determine which of these limits exist and evaluate all those that exist:

    1. \(\displaystyle\lim_{t\to 1}c(t)\)
    2. \(\displaystyle\lim_{t\to 2}c(t)\)
    3. \(\displaystyle\lim_{t\to \frac{1}{3}}c(t)\)

Answers

    1. \(\displaystyle\lim_{x\to -3}(x^3+2x+11)=-22\)
    2. \(\displaystyle\lim_{x\to 4}\frac{x^2-16}{x-4}=8\)
    3. Limit does not exist.
    1. \(\displaystyle\lim_{x\to 9}4\sqrt{x}=12\)
    2. \(\displaystyle\lim_{x\to 4}\left(11+\frac{1}{x}\right)=11+\frac{1}{4}\)
    3. \(\displaystyle\lim_{x\to 1}\frac{\cos(x-1)+2x^3}{(x+1)^2}=\frac{3}{4}\)
    4. \(\displaystyle\lim_{x\to 6}\sqrt{x^2+1}=\sqrt{37}\)
    5. \(\displaystyle\lim_{x\to 6}\frac{x-6}{x^2+36}=0\)
    6. \(\displaystyle\lim_{x\to 6}\frac{x-6}{x^2-36}=\frac{1}{12}\)
    7. \(\displaystyle\lim_{x\to 0}\frac{\sin(4x)}{10x}=\frac{2}{5}\)
    8. \(\displaystyle\lim_{x\to 0}\frac{\tan(x)}{10x}=\frac{1}{10}\)
    1. \(\displaystyle\lim_{x\to 0}10x^2\sin\left(\frac{1}{x}\right)=0;\) use squeeze theorem. First bound sine and then multiply the upper and lower bound by \(10x^2.\) Continue from there.
    2. \(\displaystyle\lim_{x\to 3^-}(x-3)f(x)=1;\) Use sequence theorem. Start with the inequality given, \(\frac{1}{x-3}\leq f(x)\leq \frac{2^{x-3}}{x-3}\) and multiply all sides of the inequality by \((x-3)\) to get \(1\leq (x-3)f(x)\leq 2^{x-3}.\) This is your upper and lower bound. Finish the problem.
    3. \(\displaystyle\lim_{x\to 0}5x^2e^{\sin\left(\frac{1}{x}\right)}=0;\) use squeeze theorem. First bound sine, which in term will bound the exponential term. Then multiply the upper and lower bound by \(10x^2.\) Continue from there.
    1. \(\displaystyle\lim_{x\to -4^-}g(x)=0\)
    2. \(\displaystyle\lim_{x\to -4^+}g(x)=-1.5\)
    3. DNE
    4. \(g(-4)=0\)
    5. No
    6. \(\displaystyle\lim_{x\to 1^-}g(x)-1\)
    7. \(\displaystyle\lim_{x\to 1^+}g(x)=1\)
    8. \(\displaystyle\lim_{x\to 1}g(x)=1\)
    9. \(\displaystyle g(1)=0\)
    10. No
    11. \(\displaystyle\lim_{x\to 6^-}g(x)=-1\)
    12. DNE
    13. DNE
    14. \(\displaystyle g(6)=-1\)
    15. No.
    1. \(\frac{1}{4}\)
    2. \(\frac{1}{4}\).
    3. Yes and the limit is equal to \(\frac{1}{4}\) because the left-hand and right-hand limit are both equal to \(\frac{1}{4}.\)

  6. \(b=\frac{77}{9}.\)

    1. \(9\)
    2. \(\infty\)
    3. DNE
    4. \(0\)
    5. \(\infty\)
    6. \(2\)
    7. \(0\)
    8. \(0\)
    1. No horizontal asymptote because \(\displaystyle\lim_{x\to\infty} f(x)=\lim_{x\to\infty}\frac{(x-2)(x+4)}{x-3}=\infty\) and \(\displaystyle\lim_{x\to-\infty} f(x)=\lim_{x\to\infty}\frac{(x-2)(x+4)}{x-3}=-\infty\). There is one vertical asymptote at \(x=3\), due to \(\displaystyle \lim_{x\to 3^-}f(x)=\displaystyle\lim_{x\to 3^{-}}\frac{(x-2)(x+4)}{x-3}=-\infty\) and \(\displaystyle \lim_{x\to 3^+}f(x)=\displaystyle\lim_{x\to 3^{-}}\frac{(x-2)(x+4)}{x-3}=\infty\)
    2. Horizontal asymptote at \(y=0\) because \(\displaystyle \lim_{x\to\infty} f(x)=\lim_{x\to\infty}\frac{(x-2)}{x(x+4)}=0\) and \(\displaystyle \lim_{x\to-\infty} f(x)=\lim_{x\to-\infty}\frac{(x-2)}{x(x+4)}=0.\) There are two vertical asymptotes. One at \(x=-4\) because \(\displaystyle \lim_{x\to -4^-} f(x)=\lim_{x\to -4^-}\frac{(x-2)}{x(x+4)}=-\infty\) and \(\displaystyle \lim_{x\to -4^+} f(x)=\lim_{x\to -4^+}\frac{(x-2)}{x(x+4)}=\infty\). The other at \(x=0\) because \(\displaystyle \lim_{x\to 0^-} f(x)=\lim_{x\to 0^-}\frac{(x-2)}{x(x+4)}=\infty\) and \(\displaystyle \lim_{x\to 0^+} f(x)=\lim_{x\to 0^+}\frac{(x-2)}{x(x+4)}=-\infty\).
    3. Horizontal asymptote at \(y=2\) because \(\displaystyle \lim_{x\to\infty} f(x)=\lim_{x\to\infty}\frac{6x-2}{3x+5}=2\) and \(\displaystyle \lim_{x\to-\infty} f(x)=\lim_{x\to-\infty}\frac{6x-2}{3x+5}=2\). There is one vertical asymptote at \(x=-\frac{5}{3}\) because \(\displaystyle\lim_{x\to-\frac{5}{3}^-}f(x)=\infty\) and \(\displaystyle\lim_{x\to-\frac{5}{3}^+}f(x)=-\infty\)
    1. \((2,\sin(1))\)
    2. \((5,0)\)
    3. \((0,\frac{5}{3}\sin\left(\frac{3}{5}\right)\)

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