without-L-Hopital-lim-x-pi-cos-x-2-x-pi-

Question Number 112782 by bemath last updated on 09/Sep/20

without L^{'} Hopital

\lim_{x \to π} \frac{\cos (\frac{x}{2})}{x - π}

Answered by john santu last updated on 09/Sep/20

s e t t i n g x = π + s, s \to 0

\lim_{s \to 0} \frac{\cos (\frac{π}{2} + \frac{s}{2})}{s} = \lim_{s \to 0} \frac{- \sin (\frac{s}{2})}{s} = - \frac{1}{2}

Answered by Dwaipayan Shikari last updated on 09/Sep/20

\lim_{x \to π} - \frac{1}{2} \frac{s i n (\frac{π}{2} - \frac{x}{2})}{(\frac{π}{2} - \frac{x}{2})} = - \frac{1}{2}

Answered by Aziztisffola last updated on 09/Sep/20

\lim_{x \to π} \frac{\cos (\frac{x}{2})}{x - π} = \cos^{'} (\frac{x}{2}) ∣_{x = π} = - \frac{1}{2} \sin (\frac{π}{2})

= - \frac{1}{2}

Commented by malwan last updated on 09/Sep/20

t h i s i s L^{^{'}} H o p i t a l

Commented by Aziztisffola last updated on 09/Sep/20

I use this property; derivitive in one

point x_{0}; this is not l^{'} Hopital rule

\lim_{x \to x_{0}} \frac{f (x) - f (x_{0})}{x - x_{0}} = f^{'} (x_{0})

let f (x) = \cos (\frac{x}{2}) a n d x_{0} = π

\lim_{x \to π} \frac{f (x) - f (π)}{x - π} = f^{'} (π) = \cos^{'} (\frac{π}{2})