4
$\begingroup$

$\lim\limits_{x\to 0} \frac{(2x \tan x)}{(1-e^x)^2}$ 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.

  • 4
    I've taken the liberty of changing that $e^2$ to $e^x$.2011-11-25

2 Answers 2

4

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.

  • 0
    Ok, ne$x$t time I will not be so hasty to give full solutions :)2011-11-25
4

${2x\tan x \over(e^x-1)^2}={2\over \cos x}\ {\sin x \over x}\Bigl({e^x -1\over x}\Bigr)^{-2}\ \to\ 2\cdot 1\cdot 1^{-2}=2\qquad (x\to0)\ .$