# Finding the $\lim\limits_{x\to 0} \frac{(1+\tan x)^{1/x} - e}{x}$

How do I find:

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

I tried L'Hospital but it does can't be applied since it's not in an indeterminate form.

Can I have some assistance? Thanks!

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As $\lim_{x\to0}(1+\tan x)^{\frac1x}=e,$ why the given expression is not in indeterminate form, just apply differentiate $(1+\tan x)^{\frac1x}$ using logarithm –  lab bhattacharjee Jun 29 '13 at 17:10
This is one of the most horrible, frustrating limits I've ever seen. I've already tried twice with different tricks (but using l'Hospital, of course) and twice I got stuck since I messed up something. One thing I can tell: it's almost sure the limits is negative and even below $\;-1.3\;$, but for that...good luck! Some sadist ideas for my first calculus course are flourishing in my mind... –  DonAntonio Jun 29 '13 at 17:26
This limit is of the form $\frac{0}{0}$, which I believe is called an indeterminate form. But as DonAntonio says, l'Hospital yields some ugly computations. Taylor would be more efficient. –  julien Jun 29 '13 at 17:27
Just apply Taylor expansion and you don't even have to think about it. L'Hospital is wasting your time. Just remember this: Power series expansions is 'pretty much' the ultimate tool to deal with analytic functions (functions with convergent power series). –  ABC Jun 29 '13 at 17:43
Recall the Taylor expansions at $0$ of $$e^u=1+u+O(u^2)\qquad \ln(1+v)=v-\frac{v^2}{2}+O(v^3)$$ and $$\tan x=x+O(x^3).$$ Thus $$\frac{\ln(1+\tan x)}{x}=\frac{1}{x}\left(\tan x-\frac{\tan^2x}{2}+O(\tan^3x)\right)$$ $$=\frac{1}{x}\left(x-\frac{x^2}{2}+O(x^3) \right)=1-\frac{x}{2}+O(x^2).$$ Then $$(1+\tan x)^\frac{1}{x}=\exp \left(1-\frac{x}{2}+O(x^2)\right)=e\exp \left(-\frac{x}{2}+O(x^2)\right)$$ $$=e\left(1-\frac{x}{2}+O(x^2)\right)=e-\frac{e}{2}x+O(x^2).$$ Finally, $$\frac{(1+\tan x)^\frac{1}{x}-e}{x}=-\frac{e}{2}+O(x)\longrightarrow -\frac{e}{2}.$$
+1. ${}{}{}{}{}{}{}{}{}$ –  Américo Tavares Jun 29 '13 at 18:19