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

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    kayte: I've edited your post (using TeX syntax). Is $x$ missing somewhere in your denominator?2011-11-25
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    My 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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    I've taken the liberty of changing that $e^2$ to $e^x$.2011-11-25

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