0 votes 0 votes $\lim_{x\rightarrow \pi } (1+\cos x)/\tan ^{2}x$ Calculus calculus engineering-mathematics + – Anmol Verma asked Dec 24, 2016 • retagged Jun 4, 2017 by Arjun Anmol Verma 344 views answer comment Share Follow See 1 comment See all 1 1 comment reply Lokesh . commented Dec 24, 2016 reply Follow Share L $= \frac{1+cos\ x}{tan^2\ x}\\= \frac{1+cos\ x}{sec^2\ x-1}\\ = \frac{1+cos\ x}{ \frac{1-cos^2\ x}{cos^2\ x}}\\ = \frac{(cos^2\ x)(1+cos\ x)}{(1-cos\ x)(1+cos\ x)}\\= \frac{cos^2\ x}{1-cos\ x}$ $\lim_{x \to \pi} L =\lim_{x \to \pi} \frac{cos^2\ x}{1-cos\ x} = \frac{cos^2\ \pi}{1-cos\ \pi}=\frac{(-1)^2}{1-(-1)} = \frac{1}{2}$ 3 votes 3 votes Please log in or register to add a comment.
Best answer 3 votes 3 votes $ \lim_{x\rightarrow \pi } (1+\cos x)/\tan ^{2}x $ applying l hospital $ \lim_{x\rightarrow \pi } (-\sin x)/2*\tan x*sec^{2}x $ it will be $\lim_{x\rightarrow \pi } (-1)/2*\sec x$ substituting ans will be 1/2 Pavan Kumar Munnam answered Dec 24, 2016 • selected Dec 24, 2016 by srestha Pavan Kumar Munnam comment Share Follow See all 0 reply Please log in or register to add a comment.