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$\lim_{x\rightarrow 0}\frac{\sin \left ( \pi \cos ^{2}x \right )}{x^{2}}$ is equal to :

Option 1)
$-\pi \;$

Option 2)
$\pi$

Option 3)
$\; \frac{\pi }{2}$

Option 4)
1

Answers (1)

best_answer

As we learnt in

Evalution of Trigonometric limit -

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

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

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

$Then\:it\:comes$

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

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

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

-

$\lim_{x\rightarrow 0}\:\:\:\frac{\sin (\pi \cos ^{2}x)}{x^{2}}$

$\lim_{x\rightarrow 0}\:\:\:\frac{\sin (\pi (1-\sin ^{2}x))}{x^{2}}$

$\lim_{x\rightarrow 0}\:\:\:\frac{\sin (\pi -\pi \sin ^{2}x)}{x^{2}}$

$\lim_{x\rightarrow 0}\:\:\:\frac{\sin (\pi \sin ^{2}x)}{x^{2}}\:\times \frac{\pi \sin^{2}x}{(\pi \sin^{2}x)}$

$= \pi$

Option 1)

$-\pi \;$

This option is incorrect.

Option 2)

$\pi$

This option is correct.

Option 3)

$\; \frac{\pi }{2}$

This option is incorrect.

Option 4)

1

This option is incorrect.

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Aadil

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