bemath-lim-x-pi-4-x-pi-cos-2-x-pi-pi-2x-cos-x-pi-2-

Question Number 106771 by bemath last updated on 07/Aug/20

@ bemath @

\lim_{x \to π} [\frac{4 (x - π) \cos^{2} x}{π (π - 2 x) \cos (x - \frac{π}{2})}] = ?

Commented by bemath last updated on 07/Aug/20

Answered by john santu last updated on 07/Aug/20

^{@ JS @}

\lim_{x \to π} [\frac{\cos^{2} x}{π (π - 2 x)}] \times \lim_{x \to π} \frac{4 (x - π)}{\cos (x - \frac{π}{2})} =

\frac{{(- 1)}^{2}}{- π^{2}} \times \lim_{x \to π} \frac{4}{- \sin (x - \frac{π}{2})} =

\frac{1}{π^{2}} \times 4 = \frac{4}{π^{2}} . (by L^{'} {Hopital}^{'} s rule)

Answered by Dwaipayan Shikari last updated on 07/Aug/20

\lim_{x \to π} \frac{4 {c o s}^{2} x}{π (π - 2 x)} \frac{(x - π)}{c o s (x - \frac{π}{2})}

\lim_{x \to π} - \frac{4}{π^{2}} \frac{x - π}{c o s (\frac{π}{2} - x)} = \lim_{x \to 0} \frac{- 4}{π^{2}} . \frac{x - π}{s i n x}

= \lim_{x \to π} \frac{4}{π^{2}} . \frac{π - x}{s i n (π - x)} = \frac{4}{π^{2}}