Chapter 5.7 Practice

Questions

Consider the function \(f\) below. Based on the graph for what \(n\) is the function \(O((x-1)^n)\):

Explain why any function that is \(O((x-x_0)^3)\) is \(o((x-x_0)^2)\).
For each function \(f\) that is given below, find the largest natural numbers \(m\) and \(n\) so that \(f\) is \(O((x-x_0)^m)\) and \(o((x-x_0)^n)\):
1. \(x_0 = -3\) and \(f(x) = (x+3)^3(x-2)^2\);
2. \(x_0 = 5\) and \(f(x) = (x-5)\sin(x-5)\);
3. \(x_0 = 3\) and \(f(x) = (x+1)^\frac{1}{5}(1-\cos(x-3))\).
For each function \(f\) that is given below, find the largest natural numbers \(m\) so that \(f\) is \(O((x-x_0)^m)\):
1. \(x_0=-6\) and \(f(x)=(x-2)^{30}(x-1)^{17}(x+6)^{20}\)
2. \(x_0=6\) and \(f(x)=(x-2)^{27}(x-6)^{35}(x-5)^{49}\)
3. \(x_0=8\) and \(f(x)=(x+2)^{98}\sin^{10}(x-8)\)
4. \(x_0=-1\) and \(f(x)=x^{87}(1-\cos(x+1))^{12}\)
5. \(x_0=3\) and \(f(x)=(x-3)^{21}(x-6)^4(x-2)^{23}(x-2)^\frac{1}{21}\)
For each function \(f\) that is given below, find the largest natural numbers \(m\) so that \(f\) is \(o((x-x_0)^m)\):
1. \(x_0=9\) and \(f(x)=(x-9)^{44}(x+6)^{44}(x-6)^{36}\);
2. \(x_0=7\) and \(f(x)=(x+9)^{14}\sin^6(x-7)\);
3. \(x_0=-8\) and \(f(x)=(x+1)^{77}(1-\cos(x+8))^{18}\);
4. \(x_0=-4\) and \(f(x)=(x+4)^{33}(x-5)^7(x+5)^{29}(x-5)^\frac{1}{3}\);
For each function \(f\) and each real number \(x_0\), write a formula for \(\Delta_{x_0}f(h)\) and simplify the expression as much as possible:
1. \(x_0 = 3\) and \(f(x) = 2x^2+1\);
2. \(x_0 = 1\) and \(f(x) = 10x^2+5x\);
3. \(x_0=1\) and \(f(x)=\cos(x)\);
4. \(x_0=1\) and \(f(x)=\frac{x}{x+1}\);
5. \(x_0=3\) and \(f(x)=\sqrt{x+6}\);
Take \(f\) and \(g\) to be the functions that are given by \[g(x) = x^2 +10\quad {\rm and}\quad f(x) = (x-11)^2.\] Perform the appropriate calculations in order to justify the following statements:
1. \(g-g(1)\) is \(\displaystyle O(x-1)\);
2. \(f\) is \(\displaystyle o(x-g(1))\);
3. \(f\circ g\) is \(\displaystyle o(x-11)\).

Answers

Based on the graph \(n=1\), or \(3\) or \(5\) or \(7\) or any odd number.
Since \(f\) is \(O((x-x_0)^3)\), then
- there is an open interval \(I\) that contains \(x_0\)
- and a bounded function \(\eta(x)\) defined on \((\mathcal{D}(f)\cap I)\setminus\{x_0\}\)
- so that for all \(x\) in \((\mathcal{D}(f)\cap I)\setminus\{x_0\}\) \(f(x)=\eta(x)(x-x_0)^3\). The claim is that \(f\) is \(o((x-x_0)^2)\). First rewrite \(f\). Instead of \(f(x)=\eta(x)(x-x_0)^3\) rewrite it as \(f(x)=[\eta(x)(x-x_0)]\cdot(x-x_0)^2,\) Define \(\eta'(x)=\eta(x)(x-x_0)\). To satisfy the definition of \(o((x-x_0^2)\), show that \(\displaystyle\lim_{x\to x_0}\eta'(x)=0\). Use the squeeze theorem. Since \(\eta\) is bounded, \[\begin{align*}m\leq &\eta(x)\leq M\\ m(x-x_0)\leq&(x-x_0)\eta(x)\leq M(x-x_0) \end{align*}\] or \(m(x-x_0)\geq(x-x_0)\eta(x)\geq M(x-x_0)\) depending on if \(x>x_0\) or \(x<x_0\). However, in either cases, both \(\displaystyle\lim_{x\to x_0}m(x-x_0)=0\) and \(\displaystyle\lim_{x\to x_0}M(x-x_0)=0\) so by the squeeze theorem \(\displaystyle\lim_{x\to x_0}\eta'(x)=0\). Thus \(f\) is \(o((x-x_0)^2)\)
1. \(O((x+3)^3)\), \(o((x+3)^2)\)
2. \(O((x-5)^2)\), \(o((x-5))\)
3. \(O((x-3))\), \(o((x-3)^0)\) or not \(o((x-3))\)
1. \(m=20\)
2. \(m=35\)
3. \(m=10\)
4. \(m=24\)
5. \(m=21\)
1. \(m=43\)
2. \(m=5\)
3. \(m=35\)
4. \(m=7\)
1. \(\Delta_{3}f(h)=2 h^2 + 12 h\)
2. \(\Delta_{1}f(h)=10h^2+25h\)
3. \(\Delta_{1}f(h)=-(1-\cos(1))\cos(h)-\sin(1)\sin(h)\)
4. \(\Delta_{1}f(h)=\frac{h}{2(h+2)}\)
5. \(\Delta_{3}f(h)=\frac{h}{\sqrt{h+9}+3}\)
1. Sketch of answer: \(g(x)-g(1)=x^2+10-11=x^2-1=(x+1)(x-1)=\eta(x)\cdot(x-1)\) where \(\eta(x)=(x+1)\).
2. Sketch of answer: \(x-g(1)=x-11\) and \(f(x)=(x-11)(x-11)=\eta(x)\cdot(x-11)\) where \(\eta(x)=(x-1)\) and \(\displaystyle\lim_{x\to 11}\eta(x)\).
3. Use the theorem from Chapter 5.7

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