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$\lim_{x\rightarrow \pi /2}\frac{\left [ 1-\tan \left ( x/2 \right ) \right ]\left [ 1-\sin x \right ]}{\left [ 1+\tan \left ( x/2 \right ) \right ]\left [ \pi -2x \right ]^{3}}$ is

Option 1)
$0$

Option 2)
$1/32$

Option 3)
$\infty$

Option 4)
$1/8$

Answers (1)

As we learnt in

Evalution of Trigonometric limit -

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

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

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

$Then\:it\:comes$

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

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

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

-

$\lim_{x \rightarrow \frac{\pi}{2}}\frac{\left [ 1- \tan \frac {x}{2}\right ] \left [ 1- \sin x\right ]}{\left [ 1+ \tan \frac{x}{2} \right ] \left [ \pi - 2x \right ]^{3}}$

$\lim_{x \rightarrow \frac{\pi}{2}}\frac{tan \left ( \frac{\pi }{4} -\frac{x}{4}\right) \left ( 1-sinx \right ) }{\left ( \pi -2x\right )^{3}}$

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

$\lim_{h \rightarrow0} \frac{tan\left ( \frac{\pi }{4} -\frac{\pi}{4} +\frac{h}{2}\right )\left ( 1- cos h\right )}{8h^{3}}$

$\lim_{h\rightarrow 0}\: \frac{tan\:h/2}{h/2\times 2} \times\frac{2sin^2h/2}{8h^2}$

$=- 1\times \frac{1}{2} \times \frac{1}{16} =\frac{1}{32}$

Option 1)

$0$

Incorrect

Option 2)

$1/32$

Correct

Option 3)

$\infty$

Incorrect

Option 4)

$1/8$

Correct

Posted by

Sabhrant Ambastha

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