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

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?

• Consider the limit of its logarithm. Commented Mar 20, 2014 at 23:35
• In order to give the answer that will help you the most, it would be good to give some background (what class is this for, what are you allowed to use, etc), or show where you are having trouble. There are many ways to answer this question; some may be too elementary and some may be too advanced.
– robjohn
Commented Mar 21, 2014 at 17:54
• @robjohn Instead, we gave him 5 answers, and hopefully there exists one that helps :)
– MT_
Commented Mar 22, 2014 at 0:15

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., $$\lim_{x\rightarrow \infty} x^{\frac{2}{x}}=1.$$

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

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

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.

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

Proof using AM-GM and Sandwich Theorem

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

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

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

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

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

• Is this something of a joke? Commented Mar 20, 2014 at 23:41
• No, not really.
– MT_
Commented Mar 21, 2014 at 0:58

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 $$\lim_{x\to\infty}x^{2/x}=1\tag{6}$$

• would the downvoter care to comment?
– robjohn
Commented Mar 22, 2014 at 17:05