How do I show that $\frac{1}{x^{2/x}} \to 1$ as $x\to\infty$?

Question

From looking at the graph, it looks like this function converges to 1 as $x\to\infty$. But with mathematical rigours, how would I show this?

Answer 1

$$\lim _{ x\rightarrow \infty }{ { x }^{ -\frac { 2 }{ x } } } =\lim _{ x\rightarrow \infty }{ { e }^{ -\frac { 2 }{ x } \ln { x } \quad \quad } } \overset { L'Hospital }{ = } \lim _{ x\rightarrow \infty }{ { e }^{ -\frac { 2 }{ x } } } =1$$

Answer 2

You can write $\frac1{x^{2/x}}$ as $e^{-2\frac{\ln x}{x}}\rightarrow 1$ as $x\rightarrow\infty$

Answer 3

$$\lim_{x\to\infty}x^{-\frac{2}{x}}=\lim_{x\to 0^+}\left(\frac{1}{x}\right)^{-2x}=\lim_{x\to 0^+}(x^x)^2=1^2=1$$ there is a topic with several proofs that $\lim_{x\to 0^+}x^x=1$

here it is: Proof of $\lim_{x \to 0^+} x^x = 1$ without using L'Hopital's rule

and another What is the value of $\lim_{x\to 0}x^x$?

Answer 4

We start by letting $u=1/x$ $$\lim _{ x\to \infty } x^{ -\frac 2x } = \lim_{u \to 0} \frac{1}{u^{2u}}$$ Write the denominator as $\left(u^u\right)^2$ and note that $\lim_{ u\to 0 } u^u =1$, which is well known.