Question

${lim}_{x\to \frac{\pi }{2}}\frac{\cot x-\cos x}{{(\pi -2x)}^3}$ equals:

• A. $\frac{1}{24}$
• B. $\frac{1}{16}$
• C. $\frac{1}{2}$
• D. $\frac{1}{4}$

Answer 1

Answer: (B)

$\frac{1}{16}$

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

$\because$ the expression is of the from $\frac{0}{0}$

so, $\underset{h\to 0}{lim}\frac{cot(\pi /2+h)-cos(\pi /2+h)}{(\pi -2(\frac{\pi }{2}+h){)}^3}$

$=\underset{h\to 0}{lim}\frac{-tanh+sinh}{8h^3}$

$=\frac{1}{8}\underset{h\to 0}{lim}[\frac{tanh}{h}\times \frac{1-cosh}{h^2}]$

$=\frac{1}{8}\underset{h\to 0}{lim}[\frac{tanh}{h}\times \frac{2si{n}^{2}h/2}{4(\frac{h}{2}{)}^2}]$

$=\frac{1}{8}[\underset{h\to 0}{lim}\frac{tanh}{h}\times \underset{h\to 0}{lim}\frac{si{n}^{2}h/2}{(h/2{)}^2}]$

$\Rightarrow \frac{1}{8\times 2}(1\times \frac{1}{2})\Rightarrow \frac{1}{8}\times \frac{1}{2}=\frac{1}{16}$

So option (b) is correct.