Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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. – 1015 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). – Mlazhinka Shung Gronzalez LeWy 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, $$ \frac{(1+\tan x)^\frac{1}{x}-e}{x}=-\frac{e}{2}+O(x)\longrightarrow -\frac{e}{2}. $$

share|cite|improve this answer
+1 Very nice . I wonder though whether the OP already covered this in her studies... – DonAntonio Jun 29 '13 at 18:23
@DonAntonio, Américo: thank you. Good question, Don. Maybe there is a trick, but right now, I can't see how to do this using L'Hospital only. – 1015 Jun 29 '13 at 18:34

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.