Question

What is the limit x->0 of (1+tan(x))^(1/x) ? I have no idea how to even start evaluating this...

Answer 1

answer is e lol

Answer 2

Note that at first glance, \(\large\displaystyle \lim_{x\to 0}(1+\tan x)^{1/x}\rightarrow 1^{\infty}\). Whenever a limit approaches this indeterminate value, we need to find a way to get this limit into the form of either \(\dfrac{0}{0}\) or \(\dfrac{\infty}{\infty}\) and then apply L'Hopital's rule.

The first step in this process is to assume that we actually get a limit. In particular, suppose that \[\large\lim_{x\to 0}(1+\tan x)^{1/x} = L\] Take natural log of both sides to get
\[\large \begin{aligned}\ln L &= \lim_{x\to 0}\ln\left[(1+\tan x)^{1/x}\right]\\ &=\lim_{x\to 0}\frac{1}{x}\cdot \ln(1+\tan x)\\ &=\lim_{x\to 0}\frac{\ln(1+\tan x)}{x}\rightarrow \frac{0}{0}\end{aligned}\] Now apply L'Hopital's rule to get \[\large \begin{aligned}\lim_{x\to 0}\frac{\ln(1+\tan x)}{x} &=\lim_{x\to 0} \frac{\dfrac{\sec^2x}{1+\tan x}}{1} \\ &= \lim_{x\to 0}\frac{\sec^2x}{1+\tan x} \\ &= \frac{1}{1+0} \\ &= 1\end{aligned}\] Thus \(\large \ln L = 1\implies L = e^1 = e\). Therefore \(\large\displaystyle \lim_{x\to 0}(1+\tan x)^{1/x} = e\).

I hope this makes sense! :-)

Answer 3

You're the best!! Thank you so much for your help.