\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)

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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