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• Notes

Indeterminate Forms and L'Hopital Rule

(1)
$\frac{0}{0} , \frac{\infty}{\infty}$
(2)
$1^{\infty} , 0^{\circ} , \infty^{0}$
(3)
$\infty-\infty , 0( \pm \infty)$
L'Hopital Rule:

(1) $F(x), g(x)$ are differentiable

(2)
$g^{\prime}(x) \neq 0$
(3)
$\lim _{x \rightarrow 9} \frac{F(x)}{g(x)}=\frac{0}{0}, \frac{\infty}{\infty}$

$\lim _{x \rightarrow a} \frac{F(x)}{g(x)}=\lim _{x \rightarrow a} \frac{F^{\prime}(x)}{g^{\prime}(x)}$
(1) L عدد حقيقي

(2)
$\pm \infty$
(2) does not exist

Q1- Find
$\lim _{x \rightarrow 1} \frac{\ln (x)}{x-1}$

$\ln (1)=0$

$\frac{d \ln (x)}{d x}=\frac{1}{x}$
(1)
$\lim _{x \rightarrow 1} \frac{\ln (x)}{x-1}=\frac{\ln (1)}{1-1}=\frac{0}{0}$
(2)
$\lim _{x \rightarrow 1} \frac{\ln (x)}{x-1}=\lim _{x \rightarrow 1} \frac{\frac{d}{d x}(\ln (x))}{\frac{d}{d x}(x-1)}=\lim _{x \rightarrow 1} \frac{\frac{1}{x}}{1}$
$=\lim _{x \rightarrow 1} \frac{1}{x}=\frac{1}{1}=1$
عدد حقيقي

Q2- Find
$\lim _{x \rightarrow \infty} \frac{e^{x}}{x^{2}}$

$e^{\infty}=\infty$

$\frac{d\left(e^{x}\right)}{d x}={1}\left(e^{x}\right)$

$=e^{x}$
(1)
$\lim _{x \rightarrow \infty}\left(\frac{e^{x}}{x^{2}}\right)=\frac{e^{\infty}}{(\infty)^{2}}=\frac{\infty}{\infty}$
(2)
$\lim _{x \rightarrow \infty} \frac{e^{x}}{x^{2}}=\lim _{x \rightarrow \infty} \frac{\frac{d}{d x}\left(e^{x}\right)}{\frac{d}{d x}\left(x^{2}\right)}=\lim _{x \rightarrow \infty} \frac{e^{x}}{2 x}=\frac{\infty}{2(x)}=\frac{\infty}{\infty}$
(3)
$\lim _{x \rightarrow \infty} \frac{e^{x}}{x^{2}}= \lim _{x \rightarrow \infty} \frac{e^{x}}{2 x}=\lim _{x \rightarrow \infty} \frac{e^{x}}{2}=\frac{e^{\infty}}{2}$
$=\frac{\infty}{2}=\infty$

Q3- Find
$\lim _{x \rightarrow 2} \frac{x^{2}-4}{x-2}$
(1)
$\lim _{x \rightarrow 2} \frac{x^{2}-4}{x-2}=\frac{(2)^{2}-4}{2-2}=\frac{4-4}{2-2}=\frac{0}{0}$
(2)
$\lim _{x \rightarrow 2} \frac{x^{2}-4}{x-2}=\lim _{x \rightarrow 2} \frac{2 x}{1}=2(2)=4$

Q4- Find
$\lim _{x \rightarrow 2} \frac{x^{3}-x^{2}-x-2}{x^{3}-3 x^{2}+3 x-2}$
(1)
$\lim _{x \rightarrow 2} \frac{x^{3}-x^{2}-x-2}{x^{3}-3 x^{2}+3 x-2}=\frac{(2)^{3}-(2)^{2}-2-2}{(2)^{3}-3(2)^{2}+3(2)-2}=\frac{0}{0}$
(2)
$\lim _{x \rightarrow 2} \frac{x^{3}-x^{2}-x-2}{x^{3}-3 x^{2}+3 x-2}=\lim _{x \rightarrow 2} \frac{3 x^{2}-2 x-1}{3 x^{2}-6 x+3}$

$=\frac{3(2)^{2}-2(2)-1}{3(2)^{2}-6(2)+3}=\frac{3(4)-4-1}{3(4)-12+3}=\frac{7}{3}$