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11.29 Limit of functions

Workshop7.pdf

  1. Prove that if \(\lim_{x \to 0} \frac{f(x)}{x} = l\) and \(b \neq 0\), then \(\lim_{x \to 0} \frac{f(bx)}{x} = bl\). What happens if \(b = 0\)?

    if \(b\neq0,\lim_{x\to0}\frac{f(bx)}{x}=\lim_{bx\to0}b\frac{f(bx)}{bx}=bl\)

    if \(b=0,\lim_{x\to0}\frac{f(bx)}{x}=\lim_{x\to0}\frac{f(0)}{x}\)

    Since \(\lim_{x\to0}\frac{f(x)}{x}=\lim_{x\to0}\frac{1}{x}\cdot\lim_{x\to0}f(x)=l\), then \(\lim_{x\to0}f(x)=0\) why?

    If \(\lim_{x\to0}f(x)=a\neq0\), then \(\lim_{x\to0}\frac{f(x)}{x}=\lim_{x\to0}\frac{1}{x}\cdot\lim_{x\to0}f(x)=\infty\)​

    If \(\lim_{x\to0}f(x)=\infty\), then \(\lim_{x\to0}\frac{f(x)}{x}=\lim_{x\to0}\frac{1}{x}\cdot\lim_{x\to0}f(x)=\infty\)

    If \(\lim_{x\to0}f(x)=-\infty\), then \(\lim_{x\to0}\frac{f(x)}{x}=\lim_{x\to0}\frac{1}{x}\cdot\lim_{x\to0}f(x)=-\infty\)

  2. Analyze \(\lim_{x \to -\infty} \frac{a_nx^n + \cdots + a_0}{b_mx^m + \cdots + b_0}\).

    If \(n>m\), \(\lim_{x\to-\infty}\frac{a_{n}x^{n}+\cdots+a_0}{b_{m}x^{m}+\cdots+b_0}=\lim_{x\to-\infty}\frac{a_{n}+\cdots+\frac{a_0}{x^{n}}}{b_{m}\frac{1}{x^{n-m}}+\cdots+\frac{b_0}{x^{n}}}=\infty\)

    1. Prove that \(\lim_{x \to 0^+} f\left(\frac{1}{x}\right) = \lim_{x \to \infty} f(x)\).

      Let \(\lim_{x\to0^{+}}f\left(\frac{1}{x}\right)\) and \(\lim_{x\to\infty}\frac{1}{x}\), then we compose: \(\lim_{x\to0^{+}}f\left(\frac{1}{x}\right)=\lim_{x\to\infty}f\left(\frac{1}{\frac{1}{x}}\right)=\lim_{x\to\infty}f\left(x\right)\) 2. Prove that \(\lim_{x \to 0^-} f\left(\frac{1}{x}\right) = \lim_{x \to -\infty} f(x)\). 3. Prove that \(\lim_{x \to \infty} f(x) = \lim_{x \to -\infty} f(-x)\).

  4. Compute the following limits:

    1. \(\lim_{x \to 1} \frac{\sin(x^2 - 1)}{x - 1}\)

      \(\lim_{x\to1}\frac{\sin(x^2-1)}{x-1}=\lim_{x\to1}\frac{\sin(x^2-1)}{x^2-1}\cdot\left(x+1\right)=1\cdot2=2\)​ 2. \(\lim_{x \to 0} \frac{\tan x}{x}\)

      \(\lim_{x\to0}\frac{\tan x}{x}=\lim_{x\to0}\frac{\sin x}{x\cos x}=1\cdot1=1\)​ 3. \(\lim_{x \to 0} \frac{x \sin x}{1 - \cos x}\)

      \(\lim_{x\to0}\frac{x\sin x}{1-\cos x}=\lim_{x\to0}x\cot\frac{x}{2}=2\lim_{x\to0}\frac{\frac{x}{2}\cos\frac{x}{2}}{\sin\frac{x}{2}}=2\cdot1\cdot\lim_{x\to0}\cos\frac{x}{2}=2\) 4. \(\lim_{x\to0}\frac{\tan^2x+2x}{x+x^2}\)

      \(\lim_{x\to0}\frac{\tan^2x+2x}{x+x^2}=\lim_{x\to0}\frac{\frac{\tan^2x}{x^2}+\frac{2}{x}}{\frac{1}{x}+1}=\frac{\lim_{x\to0}\left(\frac{\tan^2x}{x^2}+\frac{2}{x}\right)}{\lim_{x\to0}\left(\frac{1}{x}+1\right)}=\frac{1+\lim_{x\to0}\frac{2}{x}}{\lim_{x\to0}\left(\frac{1}{x}+1\right)}=\frac{\lim_{x\to0}\left(1+\frac{2}{x}\right)}{\lim_{x\to0}\left(1+\frac{1}{x}\right)}=\lim_{x\to0}\frac{x+2}{x+1}=1+\lim_{x\to0}\frac{1}{x+1}=2\)​ 5. \(\lim_{x \to 0} \frac{\sqrt{|x|}}{x}\)

      Since \(\lim_{x\to0^{+}}\frac{\sqrt{x}}{x}=\lim_{x\to0^{+}}\frac{1}{\sqrt{x}}=+\infty\) and \(\lim_{x\to0^{-}}-\frac{\sqrt{-x}}{-x}=\lim_{x\to0^{-}}-\frac{1}{\sqrt{-x}}=-\infty\), thus the limit does not exist 6. \(\lim_{x \to \infty} \frac{\sqrt{|x|}}{x}\)

      \(\lim_{x\to\infty}\frac{\sqrt{|x|}}{x}=\lim_{x\to\infty}\frac{1}{\sqrt{x}}=0\)​ 7. \(\lim_{x \to 0} \frac{\sin(ax)}{\sin(bx)}\)

      \(\lim_{x\to0}\frac{\sin(ax)}{\sin(bx)}=\lim_{x\to0}\frac{bx\sin(ax)}{ax\sin(bx)}\cdot\frac{a}{b}=\frac{a}{b}\)​ 8. \(\lim_{x \to \infty} x \left(\sqrt{x + 2} - \sqrt{x}\right)\)

      \(\lim_{x\to\infty}x\left(\sqrt{x+2}-\sqrt{x}\right)=\lim_{x\to\infty}x\left(\sqrt{x+2}-\sqrt{x}\right)\frac{\left(\sqrt{x+2}+\sqrt{x}\right)}{\left(\sqrt{x+2}+\sqrt{x}\right)}=\lim_{x\to\infty}\frac{2x}{\sqrt{x+2}+\sqrt{x}}\)​

      Since \(\frac{\sqrt{x}}{\sqrt2}=\frac{2x}{\sqrt{2x}+\sqrt{2x}}\leq\frac{2x}{\sqrt{x+2}+\sqrt{x}}\) and \(\lim\frac{\sqrt{x}}{2}\) is divergent, then \(\lim_{x \to \infty} x \left(\sqrt{x + 2} - \sqrt{x}\right)=+\infty\)

  5. Evaluate the points in the domain of \(f\) such that the function is discontinuous:

    1. \(f(x)=\begin{cases}x^2 & \text{if }x>2\\ x & \text{if }x\leq2\end{cases}\)

      \(x=2\) is discontinuous 2. \(f(x)=\begin{cases}\frac{x^2-4}{x-2} & \text{if }x\neq2\\ 1 & \text{if }x=2\end{cases}\)

      \(x=2\) is discontinuous

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