How do I solve $\lim$ as $x$ goes to infinity of $(\frac{1}{x})^{\frac{1}{x}}$ without appealing to L'Hôpital?
Note: If I take natural logs of both sides, I eventually must invoke L'Hôpital.
The best idea I've seen so far is using the Squeeze Theorem, but I have been unable to come up with functions that will squeeze $(\frac{1}{x})^{\frac{1}{x}}$.
On
Firstly; note it converging to $1$ equivalent to having $x^{1/x} \rightarrow 1$ as $x\rightarrow \infty$.
Try to imitate the proof that $(n^{1/n}) \rightarrow 1$ (as a discrete limit on the naturals). This goes as follows; it can be shown by the Mean Value Theorem, that whenever $p\geq 1$ and $y\geq 0$, $(1+y)^p \geq 1 + py$. Now define $a_n = n^{1/n} - 1$. Observe that for $n\geq 2$, $$\begin{align}\sqrt n &= (1 + a_n)^{n/2} \\ &\geq 1 + \frac{n}{2}a_n. \end{align}$$ Hence $$0 \leq a_n \leq 2\frac{\sqrt n - 1}{n},$$ so $(a_n) \rightarrow 0$ and the result follows.
What changes need to be made to this proof to show it for a continuous limit as $x\rightarrow \infty$?
I would suggest a variable change. let $y=\frac{1}{x}$
and so, switch $\lim_{x \to \infty} (\frac{1}{x})^{\frac{1}{x}}$ to $\lim_{y \to 0} y^y$
notice that $$\lim_{y \to 0} y^y = \lim_{y \to 0} e^{y \ln y}$$
since $e^x$ is continuous at $0$, we can infer the limit tends to $1$.
No L'Hopital :)