Question

$\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\frac{(cotx-cosx)}{{}_{(\mathrm{\pi }-2x)}3}$

• A. $\frac{1}{16}$
• B. $\frac{1}{8}$
• C. $\frac{1}{4}$
• D. $\frac{1}{24}$

Answer 1

The correct option is A

$\frac{1}{16}$

Explanation for the correct answer:

Step 1:

Given limit is $\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\frac{(cotx-cosx)}{{}_{(\mathrm{\pi }-2x)}3}$ solving by simplifying it we get,

$\Rightarrow \left(\frac{1}{8}\right)\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\frac{\cos x(1-\sin x)}{\sin x{\left({\displaystyle \frac{\mathrm{\pi }}{2}}-\mathrm{x}\right)}^3}$

Now, replacing $x\to \frac{\mathrm{\pi }}{2}-h$ without affecting the limit as $h\to 0$ so,

$\Rightarrow \left(\frac{1}{8}\right)\underset{h\to 0}{\lim }\frac{\cos \left({\displaystyle \frac{\mathrm{\pi }}{2}}-h\right)(1-\sin \left(\frac{\mathrm{\pi }}{2}-h\right)}{\sin \left(\frac{\mathrm{\pi }}{2}-h\right){\left({\displaystyle \frac{\mathrm{\pi }}{2}}-\left(\frac{\mathrm{\pi }}{2}-\mathrm{h}\right)\right)}^3}=\left(\frac{1}{8}\right)\underset{h\to 0}{\lim }\frac{\sin h(1-\cos h)}{\cos h{\left({\displaystyle h}\right)}^3}$

Step 2:

$\Rightarrow \left(\frac{1}{8}\right)\underset{h\to 0}{\lim }\frac{\sin h\left(2{\sin }^{2\left(\frac{h}{2}\right)}\right)}{\cos h{\left({\displaystyle h}\right)}^3}\left\{1-\cos x=2{\sin }^2\left(\frac{x}{2}\right)\right\}$

$\Rightarrow \left(\frac{1}{4}\right)\underset{h\to 0}{\lim }\frac{\sin h}{h}.{\left(\frac{\sin \left({\displaystyle \frac{h}{2}}\right)}{\left({\displaystyle \frac{h}{2}}\right)}\right)}^2.\frac{1}{\cos h}.\frac{1}{4}$

Step 3:

Now, as we know $$\begin{gathered} \underset{r\to 0}{\lim }\frac{\sin r}{r}=1 \\ \underset{r\to 0}{\lim }\cos r=1 \end{gathered}$$ so, we get

$\Rightarrow \left(\frac{1}{4}\right)\underset{h\to 0}{\lim }\frac{\sin h}{h}.{\left(\frac{\sin \left({\displaystyle \frac{h}{2}}\right)}{\left({\displaystyle \frac{h}{2}}\right)}\right)}^2.\frac{1}{\cos h}.\frac{1}{4}=\frac{1}{4}\times \frac{1}{4}=\frac{1}{16}$

So, $\underset{x\to \frac{\mathrm{\pi }}{2}}{\lim }\frac{(cotx-cosx)}{{}_{(\mathrm{\pi }-2x)}3}=\frac{1}{16}$

Hence, the correct option is (A)