Chapter 1 Limits and Derivatives(2/2)

1. $f(x) = tan x$
2. $f(x) = log_2x$

$\because \lim_{x\to n\pi + \frac{\pi}{2}^-} tanx= \infty, \lim_{x\to n\pi + \frac{\pi}{2}^+} tanx= -\infty$
$f(x) = tanx$ has vertical asymptotes $x = n\pi + \frac{\pi}{2} \forall n \in \mathbb{Z}$

$\because \lim_{x\to 0^+}log_2x = -\infty$
$f(x) = log_2x$ has vertical asymptote $x = 0$

Find $\lim_{x\to \infty} \left ( \frac{3x^2-x-2}{5x^2+4x+1} \right )$

$\lim_{x\to \infty} \left ( \frac{3x^2-x-2}{5x^2+4x+1} \cdot \frac{\frac{1}{x^2} }{\frac{1}{x^2}} \right )= \lim_{x\to \infty} \left ( \frac{3-\frac{1}{x}-\frac{2}{x^2} }{5+\frac{4}{x} +\frac{1}{x^2} } \right ) = \frac{3}{5}$
$\Rightarrow y = \frac{3}{5}$ is a horizontal asymptotes .

Find $\lim_{x\to \infty} \left ( \sqrt{x^2+1}-x \right )$

$\lim_{x\to \infty} \left ( \sqrt{x^2+1}-x \right ) = \lim_{x\to \infty} \left ( \sqrt{x^2+1}-x \right ) \cdot \frac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x} = \lim_{x\to \infty} \left ( \frac{x^2+1-x^2}{\sqrt{x^2+1}+x} \right ) = 0$
$\Rightarrow y = 0$ is a horizontal asymptote .

Find the horizontal asymptotes and vertical asymptotes of the graph of the function
$f(x) = \frac{\sqrt{2x^2+1}}{3x-5}$

$\lim_{x \to \infty}f(x) = \lim_{x\to \infty} \left ( \frac{\sqrt{2x^2+1}}{3x-5} \cdot \frac{\frac{1}{x}}{\frac{1}{x}} \right ) = \lim_{x\to \infty} \left ( \frac{\sqrt{2+\frac{1}{x^2}}}{3-\frac{5}{x}} \right ) = \frac{\sqrt{2}}{3}$

$\lim_{x \to -\infty}f(x) = \lim_{x\to -\infty} \left ( \frac{\sqrt{2x^2+1}}{3x-5} \cdot \frac{\frac{1}{x}}{\frac{1}{x}} \right ) = \lim_{x\to -\infty} \left ( \frac{\sqrt{2x^2+1}}{3x-5} \cdot \frac{-\frac{1}{|x|}}{\frac{1}{x}} \right )$
$= \lim_{x\to -\infty} \left ( \frac{\sqrt{2+\frac{1}{x^2}}}{3-\frac{5}{x}} \cdot (-1) \right ) = -\frac{\sqrt{2}}{3}$
$\therefore y = \pm \frac{\sqrt{2}}{3}$ are horizontal asymptotes of the graph of f(x)

Find the asymptote of the graph of the function
$f(x) = \frac{x^3+3}{3x^2-4x+5}$

Consider $\lim_{x\to \infty}\frac{f(x)}{x} = \lim_{x\to \infty}\frac{x^3+3}{3x^3-4x^2+5x} = \lim_{x\to \infty}\frac{1+\frac{3}{x^3}}{3-\frac{4}{x}+\frac{5}{x^2}} = \frac{1}{3}$ Let $m = \frac{1}{3}$

Consider $\lim_{x\to \infty}(f(x)-\frac{1}{3}x) = \lim_{x\to \infty} \left (\frac{x^3+3}{3x^2-4x+5} - \frac{3x^3-4x^2+5x}{3(3x^2-4x+5)} \right ) = \lim_{x\to \infty} \left (\frac{4x^2-5x+9}{9x^2-12x+15} \right ) = \frac{4}{9}$

$\therefore y = \frac{1}{3}x+\frac{4}{9}$ is a slant asymptote of the graph of f

Find the tangent line of the graph of $f(x) = e^x$ at $x=2$

$\lim_{x \to 2} \frac{f(x)-f(2)}{x-2}$ Let $h = x-2$ Then $h \to 0$ as $x \to 2$

$\because \lim_{x \to 2} \frac{f(x)-f(2)}{x-2} = \lim_{h \to 0} \frac{f(2+h)-f(2)}{h} = \lim_{h \to 0} \frac{e^{2+h}-e^2}{h}$
$= \lim_{h \to 0} \frac{e^2(e^h-1)}{h} = e^2 \lim_{h \to 0} \frac{e^h-1}{h} = e^2$
$\therefore m = e^2$, the tangent line is $y-e^2 = e^2(x-2)$

Find the tangent line of the graph of $f(x) = sinx$ at $x=\frac{\pi}{3}$

$\lim_{x \to \frac{\pi}{3}} \frac{f(x)-f(\frac{\pi}{3})}{x-\frac{\pi}{3}}$ Let $h = x-\frac{\pi}{3}$ Then $h \to 0$ as $x \to \frac{\pi}{3}$

$\because \lim_{x \to \frac{\pi}{3}} \frac{f(x)-f(\frac{\pi}{3})}{x-\frac{\pi}{3}} = \lim_{h \to 0} \frac{f(\frac{\pi}{3}+h)-f(\frac{\pi}{3})}{h} = \lim_{h \to 0} \frac{sin(\frac{\pi}{3}+h)-sin\frac{\pi}{3}}{h} = \lim_{h \to 0} \frac{sin\frac{\pi}{3}cosh+cos\frac{\pi}{3}sinh - sin\frac{\pi}{3}}{h} = \lim_{h \to 0} sin\frac{\pi}{3} \cdot \frac{cosh-1}{h} + cos \frac{\pi}{3}\frac{sinh}{h} = sin\frac{\pi}{3} \cdot 0 + cos\frac{\pi}{3} \cdot 1 = \frac{1}{2}$
$\therefore m = \frac{1}{2}$, the tangent line is $y -sin \frac{\pi}{3} = \frac{1}{2}(x-\frac{\pi}{3})$

Find $f'(0)$ if $f$ is defined by
$f(x) = \left\{\begin{matrix} \frac{1-cosx}{x} & \text{if } x \neq 0 \\ 0 & \text{if } x= 0 \end{matrix}\right.$

Consider $\lim_{h \to 0}\frac{f(0+h)-f(0)}{h} = \lim_{h \to 0}\frac{\frac{1-cosh}{h}-0}{h} = \lim_{h \to 0}\frac{1-cosh}{h^2}$
$= \lim_{h \to 0} \left ( \frac{1-cosh}{h^2} \cdot \frac{1+cosh}{1+cosh} \right ) = \lim_{h \to 0} \left ( \frac{sin^2h}{h^2} \cdot \frac{1}{1+cosh} \right ) = \frac{1}{2} \; \therefore f'(0) = \frac{1}{2}$

Find the derivative of $f(x) = \sqrt{1-x}$and compare domains (定義域) of $f$ and $f'$

Consider $\lim_{h \to 0}\frac{\sqrt{1-x-h}-\sqrt{1-x}}{h} = \lim_{h \to 0}\left ( \frac{\sqrt{1-x-h}-\sqrt{1-x}}{h} \cdot \frac{\sqrt{1-x-h}+\sqrt{1-x}}{\sqrt{1-x-h}+\sqrt{1-x}} \right ) = \lim_{h\to 0}\frac{1-x-h-1+x}{h(\sqrt{1-x-h}+\sqrt{1-x})} = \frac{-1}{2\sqrt{1-x}}$
$\Rightarrow f'(x) = \frac{-1}{2\sqrt{1-x}}$

The domoin of $f$ is $\{x| x \le 1 \} = A$
The domoin of $f'$ is $\{x| x < 1 \} = B$
$\Rightarrow A \supseteq B$

Find the derivative of f defined by $f(x) = |x|$

$f(x) = |x| = \left\{\begin{matrix} x & \text{if } x \ge 0 \\ -x & \text{if } x < 0 \end{matrix}\right.$
If $x>0$, then $\lim_{h \to 0}\frac{f(x+h)-f(x)}{h} = \lim_{h \to 0}\frac{x+h-x}{h} = 1 \Rightarrow f'(x) = 1$
If $x<0$, then
$\lim_{h \to 0}\frac{f(x+h)-f(x)}{h} = \lim_{h \to 0}\frac{-x-h+x}{h} = -1\Rightarrow f'(x) = -1$
If $x=0$, then $\begin{matrix} \lim_{h \to 0^+}\frac{f(0+h)-f(0)}{h} = \lim_{h \to 0^+}\frac{h}{h} = 1 \\ \lim_{h \to 0^-}\frac{f(0+h)-f(0)}{h} = \lim_{h \to 0^-}\frac{-h}{h} = -1 \end{matrix}$
$\Rightarrow \lim_{h \to 0}\frac{f(0+h)-f(0)}{h}$ does not exists.