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$\lim_{x\rightarrow \frac{\pi }{2}}\: \: \frac{\cot x-\cos x}{\left ( \pi -2x \right )^{3}}$ equals:

Option 1)
$\frac{1}{16}$

Option 2)
$\frac{1}{8}$

Option 3)
$\frac{1}{4}$

Option 4)
$\frac{1}{24}$

Answers (1)

best_answer

As we learnt in

Evalution of Trigonometric limit -

$\lim_{x\rightarrow a}\:\frac{sin(x-a)}{x-a}=1$

$\lim_{x\rightarrow a}\:\frac{tan(x-a)}{x-a}=1$

$put\:\:\:\:\:x=a+h\:\:\:where\:\:h\rightarrow 0$

$Then\:it\:comes$

$\lim_{h\rightarrow 0}\:\:\frac{sinh}{h}=\lim_{h\rightarrow 0}\:\:\frac{tanh}{h}=1$

$\therefore\:\:\:\lim_{x\rightarrow 0}\:\:\frac{sinx}{x}=1\:\;\;and$

$\therefore\:\:\:\lim_{x\rightarrow 0}\:\:\frac{tanx}{x}=1$

-

$\lim_{x\rightarrow \frac{\pi }{2}}\:\:\frac{cotx-cosx}{\left ( \pi -2x \right )^{3}}$

Put $x=\frac{\pi }{2}-h$

$\Rightarrow \lim_{h\rightarrow 0}\:\:\frac{cot\left ( \frac{\pi }{2} -h\right )-cos\left ( \frac{\pi }{2}-h \right )}{\left [ \pi-2 \left ( \frac{\pi }{2}-h \right ) \right ]^{3}}$

$\Rightarrow \lim_{h\rightarrow 0}\:\:\frac{tanh-Sinh}{\left ( \pi -\pi +2h \right )^{3}}$

$\Rightarrow \lim_{h\rightarrow 0}\:\:\frac{tanh\left ( 1-cosh \right )}{8h^{3}}$

$\Rightarrow \frac{1}{8}\lim_{h\rightarrow 0}\:\:\frac{tanh}{h}\cdot \frac{2sin^{2}\frac{h}{2}}{h^{2}}$

$\Rightarrow \frac{1}{8}\times 1\times \frac{1}{2}=\frac{1}{16}$

Option 1)

$\frac{1}{16}$

This option is correct.

Option 2)

$\frac{1}{8}$

This option is incorrect.

Option 3)

$\frac{1}{4}$

This option is incorrect.

Option 4)

$\frac{1}{24}$

This option is incorrect.

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prateek

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