lim-x-pi-cos-x-sin-2x-1-x-2-pi-2-A-1-2pi-C-1-B-1-pi-D-Does-Not-exist-

Question Number 187270 by MikeH last updated on 15/Feb/23

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

A . \frac{1}{2 π} C . 1

B . \frac{1}{π} D . Does Not exist

Answered by horsebrand11 last updated on 15/Feb/23

\lim_{x \to π} \frac{1 - \cos (π - x) - \sin 2 (π - x)}{(x + π) (x - π)}

= \frac{1}{2 π} . \lim_{x \to 0} \frac{1 - \cos x - \sin 2 x}{- x}

= - \frac{1}{2 π} [\lim_{x \to 0} \frac{1 - \cos x}{x} - \lim_{x \to 0} \frac{\sin 2 x}{x}]

= - \frac{1}{2 π} [0 - 2] = \frac{1}{π}

Commented by MikeH last updated on 16/Feb/23

thanks

Answered by Frix last updated on 15/Feb/23

Simply use l^{'} H \hat{o p i t a l} \Rightarrow answer is \frac{1}{π}

Answered by CElcedricjunior last updated on 15/Feb/23

\lim_{x \to π} \frac{c o s x + s i n (2 x) + 1}{x^{2} - π^{2}} = \frac{0}{0} = F I M o i v r e

\lim_{x \to π} \frac{- s i n x + 2 c o s (2 x)}{2 x} = \frac{2}{2 π} = \frac{1}{π} ★ c e d r i c j u n i o r

B) \frac{1}{π}