\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

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