Finding $\lim_{x\to 0} \frac{(1+\tan x)^{\frac{1}{x}}-e}{x}$

Solution 1:

You may write, for $x$ near $0$, $$ \tan x=x+\frac{x^3}{3}+\mathcal{O}(x^5) $$ $$ \log(1+\tan x)=x-\frac{x^2}{2}+\mathcal{O}(x^3) $$ $$ \frac1x\log(1+\tan x)=1-\frac{x}{2}+\mathcal{O}(x^2) $$ then $$ e^{\frac1x\log(1+\tan x)}=e^{1-\frac{x}{2}+\mathcal{O}(x^2)}=e(1-\frac{x}{2}+\mathcal{O}(x^2)) $$ and $$ \frac{(1+\tan x)^{\frac{1}{x}}-e}{x}=\frac{e(1-\frac{x}{2}+\mathcal{O}(x^2))-e}{x}=-\frac{e}{2}+\mathcal{O}(x) $$ giving $\displaystyle -\frac{e}{2} $ as limit.

Solution 2:

You almost have the idea. I will just hint that instead of letting $y$ be $\lim_{x \to 0} (1+\tan x)^{1/x}$, let it be the expression that you want to get the limit of. That is, let $y = \frac{(1+\tan x)^{1/x} - e}{x}$. Then take the natural log before taking the limit.