Difficult power limit $\lim_{x \to \infty} x^{\left ( -\frac{1}{x} \right )}$

Question

This is my work so far

$$\lim_{x \to \infty} x^{\left ( -\frac{1}{x} \right )}= \lim_{x \to \infty}\frac{1}{x^\left({\frac{1}{x}} \right)}=\lim_{x \to \infty}\frac{1}{e^{\ln x^\left({\frac{1}{x}}\right)}}=\lim_{x \to \infty}\frac{1}{e^{\frac{\ln x}{x}}}$$

Now what's the next step?

I'm stuck and I don't know what to do! I need to solve this without differentiation! Can someone help me?

Answer 1

The proof can be performed without knowledge of the $\log$ function.

Let $2\le n\le x\le n+1.$ Then $$1\le x^{1/x} \le (n+1)^{1/n}\le (2n)^{1/n}=:1+y_n$$ Hence by the binomial formula we obtain $$2n=(1+y_n)^{n}\ge {n\choose 2}y_n^2$$ which gives $$y_n^2\le {4\over n-1}\le {4\over x-2 }$$ Summarizing we get for $x\ge 2$ $$1 \le x^{1/x}\le 1+{2\over \sqrt{x-2}}$$ Therefore the limit is equal $1.$

Answer 2

Your work is fine and, as noticed in the comments, you reduce all to the well known limit

$$\lim_{x \to \infty}\frac{\ln x}{x}=0$$

which can be proved without l'Hospital's rule in many different ways as already shown for example here

• Evaluation of $ \lim_{x\rightarrow \infty}\frac{\ln (x)}{x}$ using Squeeze theorem
• Find $\lim_{x\to \infty}\frac{\ln x}{x}=0$ without l'Hopital's rule
• Proof that limit $\frac{\ln(x)}x$ for $x\to\infty$ is $0$.