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###### ${selected_topic_name} • Notes • Comments & Questions Q5- Find $\lim _{x \rightarrow \pi^{-}} \frac{\sin x}{1-\cos x}$ $\frac{d \sin x}{d x}=\cos x$ $\frac{d \cos x}{d x}=-\sin x$ $\lim _{x \rightarrow \pi^{-}} \frac{\sin x}{1-\cos x}=\lim _{x \rightarrow \pi^{-}} \frac{\cos x}{-(-\sin x)}=\frac{\cos (\pi)}{\sin (\pi)}=\frac{-1}{0}=\infty$ (1) $\lim _{x \rightarrow \pi^{-}} \frac{\sin (x)}{1-\cos (x)}=\frac{\sin (\pi)}{1-\cos (\pi)}=\frac{0}{1-(-1)}=\frac{0}{1+1}$ $=\frac{0}{2}=0$ Q6- Find $\lim _{x \rightarrow 0} \frac{\tan x-x}{x^{3}}$ $\tan (0)=0$ $\sec (0)=1$ $\frac{d \tan (x)}{d x}=\sec ^{2}(x)$ $\frac{d}{d x} \sec ^{2}(x)=2 \sec ^{2} x \tan x$ (1) $\lim _{x \rightarrow 0} \frac{\tan x-x}{x^{3}}=\frac{\tan (0)-0}{0^{3}}= \frac{0-0}{0}=\frac{0}{0}$$\$

(2)
$\lim _{x \rightarrow 0} \frac{\tan x-x}{x^{3}}=\lim _{x \rightarrow 0} \frac{\sec ^{2} x-1}{3 x^{2}}$

$=\frac{\sec ^{2}(0)-1}{3(0)^{2}}=\frac{1-1}{0}=\frac{0}{0}$

(3)
$\lim _{x \rightarrow 0} \frac{\tan x-x}{x^{3}}=\lim _{x \rightarrow 0} \frac{\sec ^{2} x-1}{3 x^{2}}=\lim _{x \rightarrow 0} \frac{2 \sec ^{2} x \tan x}{6 x}$

$=\frac{1}{3} \lim \frac{\sec ^{2} x \tan x}{x}$

$=\frac{1}{3} \lim _{x \rightarrow 0}\left(\sec ^{2} x\right) . \lim _{x \rightarrow 0}\left(\frac{\tan x}{x}\right)$

$\lim _{x \rightarrow 0} \sec ^{2}(x)=\sec ^{2}(0)=1$

$\lim _{x \rightarrow 0} \frac{\tan x}{x}=\frac{\tan \theta}{0}$

$=\frac{0}{0}$

$=\frac{1}{3} \cdot(1) \cdot \lim _{x \rightarrow 0} \frac{\sec ^{2} x}{1}=\frac{1}{3} \cdot \sec ^{2}(0)$

$=\frac{1}{3} \cdot(1)=\frac{1}{3}$

Q7- Find
$\lim _{x \rightarrow 0^{+}} x \ln (x)$

$\ln \left(0^{+}\right)=-\infty$

$\frac{d \ln (x)}{d x}=\frac{1}{x}$

(1)
$\lim _{x \rightarrow 0^{+}} x \ln (x)=(0) \ln (0)=0 \cdot(-\infty)$

(2)
$\lim _{x \rightarrow 0^{+}} x \ln (x)=\lim _{x \rightarrow 0^{+}} \frac{\ln (x)}{\frac{1}{x}}=\frac{\ln (0)}{\frac{1}{0}}=\frac{-\infty}{\infty}$

(1)
$\frac{0}{0} , \frac{\infty}{\infty}$

(2)
$1^{\infty} , 0^{0} , \infty^{0}$

$\infty-\infty,0(-\infty)$

$\frac{a}{b}$

$\lim _{x \rightarrow 0^{+}} \frac{\ln (x)}{\frac{1}{x}}=\lim _{x \rightarrow 0^{+}} \frac{\frac{1}{x}}{\frac{-1}{x^{2}}}=\lim _{x \rightarrow 0^{+}} \frac{-x^{2}}{x}=\lim _{x \rightarrow 0^{+}}(-x)$

$=-(0)=0$

Q8- Find
$\lim _{x \rightarrow \pi} \frac{\sin x}{\sqrt{x-\pi}}$

$\sin (\pi)=0$

$\frac{d \sin (x)}{d x}=\cos x$

(1)
$\lim _{x \rightarrow \pi^{+}} \frac{\sin x}{\sqrt{x-\pi}}=\frac{\sin (\pi)}{\sqrt{\pi-\pi}}=\frac{0}{\sqrt{0}}=\frac{0}{0}$

(2)
$\lim _{x \rightarrow \pi^{+}} \frac{\sin x}{\sqrt{x-\pi}}=\lim _{x \rightarrow \pi^{+}} \frac{\cos (x)}{\frac{1}{2 \sqrt{x_{-} \pi}}}=\lim _{x \rightarrow \pi^{+}} \frac{\cos (x)}{\frac{1}{2 \cdot(x-\pi)^{\frac{1}{2}}}}$

$\begin{array}{l}{=\lim _{x \rightarrow \pi^{+}} 2 \cdot \cos x \cdot(x-\pi)^{\frac{1}{2}}=2 \cos (\pi) \cdot(\pi-\pi)^{\frac{1}{2}}}\end{array}$

$=0$

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