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• Notes

Q13-Find
$\lim _{x \rightarrow 0}({scc} (x)-\cot (x) )$

${cot x}=\frac{\cos x}{\sin x}$

${csc}(x)=\frac{1}{\sin x}$

(1)
$\lim _{x \rightarrow 0}(\csc (x)-\cot (x))=\csc (0)-\cot (0)$

$=\frac{1}{\sin (0)}-\frac{\cos (0)}{\sin (0)}=\frac{1}{0}-\frac{1}{0}$

$=\infty-\infty$

$\lim _{x \rightarrow 0}(\csc (x)-\cot x)=\lim _{x \rightarrow 0}\left(\frac{1}{\sin x}-\frac{\cos x}{\sin x}\right)=\lim _{x \rightarrow 0} \left(\frac{1-\cos x}{\sin x}\right)$

$=\lim _{x \rightarrow 0}\left(\frac{-(-\sin x)}{\cos x}\right)=\lim _{x \rightarrow 0}\left(\frac{\sin x}{\cos x}\right)=\frac{\sin (0)}{\cos (0)}=\frac{0}{1}=0$

Q14- Find
$\lim _{x \rightarrow \infty} x^{2}\left(1-\cos \left(\frac{1}{x}\right)\right)$

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

(1)
$\lim _{x \rightarrow \infty} x^{2}\left(1-\cos \frac{1}{x}\right)=(\infty)^{2}\left(1-\cos \left(\frac{1}{\infty}\right)\right)=\infty(1-\cos (0))$

$=\infty(1-1)=\infty(0)$

(2)
$\lim _{x \rightarrow \infty} x^{2}\left(1-\cos \left(\frac{1}{x}\right)\right)=\lim _{x \rightarrow \infty}\left(\frac{1-\cos \left(\frac{1}{x}\right)}{x^{-2}}\right)=\lim _{x \rightarrow \infty}\left(\frac{-\left(-\left(-\frac{1}{x^{2}}\right) \sin \left(\frac{1}{x}\right)\right)}{-2 x^{-3}}\right)$

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

$=\frac{1}{2} \lim \left(\frac{\sin \left(\frac{1}{x}\right)}{x^{-3} \cdot x^{2}}\right)=\frac{1}{2} \lim \left(\frac{\sin \frac{1}{x}}{\frac{1}{x}}\right)$

$=\frac{1}{2}\left(\frac{\sin \left(\frac{1}{\infty}\right)}{\frac{1}{\infty}}\right)=\frac{1}{2}\left(\frac{\sin (0)}{0}\right)=\frac{1}{2}\left(\frac{0}{0}\right)$

نشتق مرة ثانية
$=\frac{1}{2} \lim \left(\frac{-\frac{1}{x_{2}} \cdot \cos \frac{1}{x}}{-\frac{1}{x^{2}}}\right)$

$=\frac{1}{2} \lim _{x \rightarrow \infty}\left(\cos \frac{1}{x}\right)=\frac{1}{2} \cdot \cos \frac{1}{\infty}$

$=\frac{1}{2} \cos (0)=\frac{1}{2} \cdot(1)=\frac{1}{2}$

Q15- Find b
$\lim _{x \rightarrow 0} \frac{\sin (3 x)-3 x+b x^{3}}{x^{3}}=0$

(1)
$\lim _{x \rightarrow 0}\left(\frac{\sin 3 x-3 x+6 x^{3}}{x^{3}}\right)$

$=\lim \left(\frac{\sin (3(0))-3(0)+b(0)^{3}}{(0)^{3}}\right)$

$=\frac{0-0+0}{0}=\frac{0}{0}$

(2)
$\lim _{x \rightarrow 0}\left(\frac{\sin 3 x-3 x+b x^{3}}{x^{3}}\right)=\lim _{x \rightarrow 0}\left(\frac{3 \cdot \cos 3 x-3+3 b x^{2}}{3 x^{2}}\right)$

$\frac{d}{d x} \cos x=$

$=-3 \sin 3 x$

$=\lim _{x \rightarrow 0}\left(\frac{3 \cdot-3 \sin 3 x+6b x}{6 x}\right)$

$=\lim _{x \rightarrow 0}\left(\frac{-9.3 \cos 3 x+6 b}{6}\right)=\lim _{x \rightarrow 0}\left(\frac{-27 \cos 3 x+6 b}{6}\right)$

$=\frac{-27 \cos (0)+6 b}{6}=\frac{-27(1)+6 b}{6}=\frac{-27+6 b}{6}=0$

$\frac{-27+6 b}{6}=0 \Rightarrow-27+6 b=0 \Rightarrow 6b = +27$

$\frac{6 b}{6}=+\frac{27}{6}$

$b=\frac{+27}{6}$

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