Question

please ans this question I'm simple and clear manner ​

Answer 1

Answer:

$\tt{ \lim_{x \to \dfrac{\pi}{2}} \dfrac{cot\,x - cos\,x}{(\dfrac{\pi}{2}-x)^{3}} = \dfrac{1}{2}}$

Step-by-step explanation:

Given: $\tt{ \lim_{x \to \dfrac{\pi}{2}} \dfrac{cot\,x - cos\,x}{(\dfrac{\pi}{2}-x)^{3}} }$

$\tt{ = \lim_{x \to \dfrac{\pi}{2}} \dfrac{\dfrac{cos\,x}{sin\,x} - cos\,x}{(\dfrac{\pi}{2}-x)^{3}} }$

$\tt{ = \lim_{x \to \dfrac{\pi}{2}} \dfrac{cos\,x - sin\,x \ cos\,x}{sin\,x \ (\dfrac{\pi}{2}-x)^{3}} }$

$\tt{ = \lim_{x \to \dfrac{\pi}{2}} \dfrac{cos\,x \ (1 - sin\,x)}{sin\,x \ (\dfrac{\pi}{2}-x)^{3}} }$

$\tt{ = \lim_{x \to \dfrac{\pi}{2}} \dfrac{sin\,(\dfrac{\pi}{2}-x) \ (1 - sin\,x)}{sin\,x \ (\dfrac{\pi}{2}-x)(\dfrac{\pi}{2}-x)^{2}} }$

$\tt{ = \lim_{x \to \dfrac{\pi}{2}} \dfrac{1 - sin\,x}{sin\,x \ (\dfrac{\pi}{2}-x)^{2}} }$

$\tt{ = \lim_{x \to \dfrac{\pi}{2}} \dfrac{cosec\,x - 1}{(\dfrac{\pi}{2}-x)^{2}} }$

On applying L'Hopital's Rule, we get

$\tt{ = \lim_{x \to \dfrac{\pi}{2}} \dfrac{-cosec\,x \ cot\,x}{-2(\dfrac{\pi}{2}-x)} }$

$\tt{ = \lim_{x \to \dfrac{\pi}{2}} \dfrac{\dfrac{1}{sin\,x} \times \dfrac{cos\,x}{sin\,x}}{2(\dfrac{\pi}{2}-x)} }$

$\tt{ = \lim_{x \to \dfrac{\pi}{2}} \dfrac{cos\,x}{2\,sin^{2}\,x(\dfrac{\pi}{2}-x)} }$

$\tt{ = \lim_{x \to \dfrac{\pi}{2}} \dfrac{sin\,(\dfrac{\pi}{2}-x)}{2\,sin^{2}\,x \ (\dfrac{\pi}{2}-x)} }$

$\tt{ = \lim_{x \to \dfrac{\pi}{2}} \dfrac{1}{2\,sin^{2}\,x} }$

$\tt{ = \dfrac{1}{2\,sin^{2}(\dfrac{\pi}{2})} }$

= 1/2