# Lim. x (e2+x-e2)/1-cosxx=> 0

9 years ago
Lim. x (e2+x-e2)/1-cosx
x=> 0
it is a 0/0 form,

so differentiate above and below,
Lim. (x (e2+x)+ (e2+x-e2))/sinx
x=> 0

again differentiate,

Lim. (x (e2+x)+e2+x+ (e2+x))/cosx
x=> 0

(1+1)/1=2

Arun Kumar IIT Delhi
9 years ago
Hi

$\\\lim _{x\to \:0}\left(\frac{x\left(e^{2+x}-e^x\right)}{1-\cos \left(x\right)}\right) \\\mathrm{Concept\:}\lim _{x\to a^-}f\left(x\right)\ne \lim _{x\to a^+}f\left(x\right)\mathrm{\:then\:the\:limit\:does\:not\:exist} \\\lim _{x\to \:0+}\left(\frac{x\left(e^{2+x}-e^x\right)}{1-\cos \left(x\right)}\right) \\\lim _{x\to a}[\frac{f\left(x\right)}{g\left(x\right)}]=\frac{\lim _{x\to a}f\left(x\right)}{\lim _{x\to a}g\left(x\right)}\mathrm{,\:where\:}\lim _{x\to a}g\left(x\right)\ne 0 \\=\frac{\lim _{x\to \:0+}\left(\left(e^2-1\right)e^x\left(x+1\right)\right)}{\lim _{x\to \:0+}\left(\sin \left(x\right)\right)} \\=\frac{e^2-1}{0} \\=\infty \:$

$\\\lim _{x\to \:0-}\left(\frac{x\left(e^{2+x}-e^x\right)}{1-\cos \left(x\right)}\right) \\=\lim _{x\to \:0-}\left(\frac{\left(e^2-1\right)e^x\left(x+1\right)}{\sin \left(x\right)}\right) \\=\frac{\lim _{x\to \:0-}\left(\left(e^2-1\right)e^x\left(x+1\right)\right)}{\lim _{x\to \:0-}\left(\sin \left(x\right)\right)} \\=\frac{e^2-1}{0} \\=-\infty \:$
because it approaches from the negative side.
So the limit diverges

Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty