$$\lim\limits_{x\to 0} \frac{(2x \tan x)}{(1-e^x)^2}$$ 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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0kayte: I've edited your post (using TeX syntax). Is $x$ missing somewhere in your denominator? – 2011-11-25
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2My guess is that the denominator should be $(1 - e^{\color{Red}{x}})^2$ instead of what's written. Then you will get the $0/0$ indeterminate form as you expect. – 2011-11-25
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4I've taken the liberty of changing that $e^2$ to $e^x$. – 2011-11-25