$$\lim\limits_{x\to 0} \frac{(2x \tan x)}{(1-e^x)^2}$$ 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? – Martin Sleziak Nov 25 '11 at 09:31
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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. – Srivatsan Nov 25 '11 at 09:31
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4I've taken the liberty of changing that $e^2$ to $e^x$. – Gerry Myerson Nov 25 '11 at 09:52
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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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1I would have been interested in seeing how far kayte had gotten, before posting a complete solution. – Gerry Myerson Nov 25 '11 at 09:54
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1@GerryMyerson Actually I didn't put the derivatives into my answer to let the real work be up to him. The essential "trick" is to apply L'Hôpital twice, but in my opinion you cannot really tell him that without giving my answer. – Listing Nov 25 '11 at 09:59
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2In my opinion, you can let OP tell you what he (or she - I've been assuming kayte is a woman's name) has done on the problem, where he got stuck. Then you're in a better position to give encouragement and maybe give the one little missing piece that unlocks the puzzle. I don't mean for this to be taken as criticism of what you have done, it's just meant to suggest another possible way of doing things. – Gerry Myerson Nov 25 '11 at 10:36
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$${2x\tan x \over(e^x-1)^2}={2\over \cos x}\ {\sin x \over x}\Bigl({e^x -1\over x}\Bigr)^{-2}\ \to\ 2\cdot 1\cdot 1^{-2}=2\qquad (x\to0)\ .$$
Christian Blatter
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