$\lim\limits_{x\to 0} \frac{(2x \tan x)}{(1-e^x)^2}$ I have tried using l'hospital's rule to solve for 0/0 indeterminations, but with no success. Can anyone help me to find the solution to this problem? Thank you in advance for any advice.
Can't get out of a $\frac 00$ indeterminate form for this limit
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4I've taken the liberty of changing that $e^2$ to $e^x$. – 2011-11-25
2 Answers
Set $g(x)=2 x \tan (x)$, $h(x)=\left(1-e^x\right)^2$, $f(x)=\frac{g(x)}{h(x)}$
Then as $\lim \limits_{x \to 0}g(x)=\lim\limits_{x \to 0}h(x)=0$ you have that
\lim_{x \rightarrow 0}\frac{g(x)}{h(x)}=\lim_{x \rightarrow 0}\frac{g'(x)}{h'(x)}
if the right hand side exists by l'Hôpital. But as \lim\limits_{x \to 0}g'(x)=\lim\limits_{x \to 0}h'(x)=0 you have that
\lim\limits_{x \to 0}\frac{g'(x)}{h'(x)}=\lim\limits_{x \to 0}\frac{g''(x)}{h''(x)}
if the right hand side exists by l'Hôpital. But by evaluating this you can see that
\lim_{x \rightarrow 0}\frac{g''(x)}{h''(x)}=\frac{4}{2}=2
And therefore you have your limit.
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0Ok, ne$x$t time I will not be so hasty to give full solutions :) – 2011-11-25
${2x\tan x \over(e^x-1)^2}={2\over \cos x}\ {\sin x \over x}\Bigl({e^x -1\over x}\Bigr)^{-2}\ \to\ 2\cdot 1\cdot 1^{-2}=2\qquad (x\to0)\ .$