According to Wolfram Alpha, this limit evaluates to 1.
However, I don't know how to prove this fact.
My problems are with $\tan(\frac{1}{x})$. As I understand, when $x \to 0$, $\tan(\frac{1}{x})$ is not defined, because $\tan(\frac{1}{x}) = \frac{\sin(\frac{1}{x})}{\cos(\frac{1}{x})}$ and both the $\sin$ and the $\cos$ oscillate as they approach infinity, hence there is no definite value for them.
Maybe this problem requires an approach that I am not familiar with, but it appears as homework in a Calculus I course, so it should not involve super advanced maths techniques.
Thanks for your help!
Best Answer
The limit does not exist.
Note that there are arbitrarily large positive solutions to the equation $\tan \theta=\theta$ (sketch the graphs: tan regularly goes from $0$ to $+\infty$ and there is a solution in each such segment). For every solution, setting $x=1/\theta$ gives $(1+x)^{\tan(1/x)}=(1+1/\theta)^\theta\geq 2$.
This gives an infinite sequence approaching $0$ from above for which the function exceeds $2$; you can easily find a sequence for which the function takes the value $1$ (e.g. $x_n=1/(n\pi)$), so there is no limit.