Why is $ \lim_{x \to \infty} \ x^{2/x} = 1$

Question

Why is $\displaystyle \lim_{x \to \infty} \ x^{2/x} = 1$ since this is an indeterminate form $\infty^{0}$ and I can't see any manipulation that would suggest this result?

Answer 1

If $f(x)=x^{\frac{2}{x}}$, then $\ln f(x)=\ln x^{\frac{2}{x}}=\frac{2}{x}\ln x$, and $$\lim_{x\rightarrow \infty}\ln f(x)=\lim_{x\rightarrow \infty}\frac{2}{x}\ln x =\lim_{x\rightarrow \infty}\frac{2\ln x}{x},$$ and the indetermination now is in the form $\frac{\infty}{\infty}$. Then, for L'Hospital rule: $$\lim_{x\rightarrow \infty}\ln f(x)=\lim_{x\rightarrow \infty}\frac{2\ln x}{x}=\lim_{x\rightarrow \infty} \frac{2\frac{1}{x}}{1}=0,$$ then $$0=\lim_{x\rightarrow \infty}\ln f(x)=\ln \lim_{x\rightarrow \infty} f(x). $$ Now, $\ln w=0\Leftrightarrow w=1$, then $$ \lim_{x\rightarrow \infty} f(x)=1,$$ i.e., $$ \lim_{x\rightarrow \infty} x^{\frac{2}{x}}=1.$$

Answer 2

$$x^{2/x} = (e^{\log(x)})^{2/x} = \exp(\frac{2 \log(x)}{x}) \rightarrow \exp(0) = 1 \quad \textrm{as} \quad x \to \infty$$

where in the limit I use the fact that $x$ dominates $\log(x)$ as $x \to \infty$.

Answer 3

Let $ y=x^{2/x} $. Then, $\ln y=\dfrac {2\ln x}{x} $. Take the limit as $ x $ goes to infinity. Then, finally, take the exponential ($ e^\cdot $) of your resulting answer. This is the value of your limit.

Answer 4

$\lim \limits_{x \to \infty} x^{\frac{1}{x}} = 1$

Proof using AM-GM and Sandwich Theorem

$\frac{1 + 1 + 1 + \dots + \sqrt{x} + \sqrt{x}}{x} \geq \sqrt[x]{x} \geq 1$

$\frac{x - 2 + 2\sqrt{x}}{x} \geq \sqrt[x]{x} \geq 1$

$1 - \frac{2}{x} + \frac{2}{\sqrt{x}} \geq \sqrt[x]{x} \geq 1$

$\lim \limits_{x \to \infty} 1 - \frac{2}{x} + \frac{2}{\sqrt{x}} = 1$

$\therefore \lim \limits_{x \to \infty} x^{\frac{1}{x}} = 1 \implies \lim \limits_{x \to \infty} x^\frac{2}{x} = 1$

Answer 5

For all $x\in\mathbb{R}$: $1+x\le e^x$. Substitute $x\mapsto\frac xn$ and then raise to the $2/x$ power (assuming $x\gt0$): $$ \left(\frac xn\right)^{2/x}\le\left(1+\frac xn\right)^{2/x}\le e^{2/n}\tag{1} $$ Multiplying by $n^{2/x}$ gives the following for any $x$ and $n$ greater than $0$: $$ x^{2/x}\le e^{2/n}n^{2/x}\tag{2} $$ Thus, $$ \begin{align} \lim_{x\to\infty}x^{2/x} &\le e^{2/n}\lim_{x\to\infty}n^{2/x}\\ &=e^{2/n}\tag{3} \end{align} $$ Since $(3)$ is true for any $n$, we must have that $$ \lim_{x\to\infty}x^{2/x}\le1\tag{4} $$ For $x\ge1$, we have $x^{2/x}\ge1$; therefore, $$ \lim_{x\to\infty}x^{2/x}\ge1\tag{5} $$ Inequalities $(4)$ and $(5)$ yield $$ \lim_{x\to\infty}x^{2/x}=1\tag{6} $$