Question

$\underset{x\to \frac{\pi }{2}}{lim}\frac{\cot x-\cos x}{{\left(\pi -2x\right)}^3}$

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

Answer 1

The correct option is B $\frac{1}{16}$

Given $\underset{x\to \frac{\pi }{2}}{lim}\frac{\cot x-\cos x}{{(\pi -2x)}^3}$

$l=\underset{x\to \frac{\pi }{2}}{lim}\frac{-{\csc }^{2}x+\sin x}{3{(\pi -2x)}^2(-2)}$ [L-Hospital rule]

$l=\underset{x\to \frac{\pi }{2}}{lim}\frac{\cos x-2\csc x(-\csc x\cot x)}{6(\pi -2x)(-2)(-2)}$ [again L-Hospital rule]

$l=\underset{x\to \frac{\pi }{2}}{lim}\frac{-\sin x+2\cot x\left(2\csc x(-\csc x)\left(\cot x\right)\right)+2{\csc }^{2}x.(-{\csc }^{2}x)}{6(-2)(-2)(-2)}$ (again L-Hospital applied)

$l=\underset{x\to \frac{\pi }{2}}{lim}\frac{-\sin x-2{\csc }^{2}x(2{\cot }^{2}x+{\csc }^{2}x)}{6(-2)(-2)(-2)}$

$l=\frac{1}{-48}(-1-2(0+1\left)\right)$

$l=\frac{-3}{-48}$

$l=\frac{1}{16}$