$$\lim _{x\to \infty }x^{\frac{1}{x}}$$ How can I solve this? I want the most simple way to do it. Is there a nice log rule I can use here? I'm definitely not at the stage of using the limit chain rule as suggested on SymboLab.
I couldn't get anywhere with it using the methods I know ($0/0, a/0, a/\infty,$ basic logs).
Edit: This is an extremely snaky move from me to add the limits-without-lhopital tag now. Is it too late, or do I need to add a separate question?
firstly,write it as $x^{ \frac { 1 }{ x } }={ e }^{ \frac { 1 }{ x } \ln { x } }$ now we have such limit $$\lim _{ x\to \infty } e^{ \frac { \ln { x } }{ x } }=\lim _{ x\rightarrow \infty }{ { e }^{ \frac { f\left( x \right) \rightarrow \infty }{ g\left( x \right) \rightarrow \infty } } } ,\quad where\quad f\left( x \right) =\ln { x } ,g\left( x \right) =x $$ so we can apply L'hospital's rule
$$\lim _{ x\to \infty } x^{ \frac { 1 }{ x } }=\lim _{ x\to \infty } e^{ \frac { 1 }{ x } \ln { x } }=\lim _{ x\to \infty } e^{ \frac { 1 }{ x } }=1$$
If you know that for any $\;\epsilon>0\;$ we have that $\;\log x<x^\epsilon\;$ for any $\;0<x\in\Bbb R\;$ big enough , you then don't need l'Hospital since we can write
$$\frac{\log x}x<\frac{x^\epsilon}x=\frac1{x^{1-\epsilon}}\xrightarrow[x\to\infty]{}0\;,\;\;\text{for say}\;\;0<\epsilon<1$$
Using that $\;\log x>\log1=0\;$ for $\;x>1\;$, use the squeeze theorem to deduce that
$$\lim_{x\to\infty}\frac{\log x}x=0$$
and, as in the other answers, use now the algebraic equality
$$x^{1/x}=e^{\frac1x\log x}\implies\lim_{x\to\infty}x^{1/x}=e^{\lim\limits_{x\to\infty}\frac1x\log x}=e^0=1$$
$x^{1/x}=e^{\ln(x)/x}$ so the limit is $\lim\limits_{x\rightarrow +\infty}e^{\ln(x)/x}$.
You know that $\lim\limits_{+\infty}\ln(x)/x=0$ this implies that $\lim\limits_{x\rightarrow +\infty}e^{\ln(x)/x}=1.$
If you consider this for integer $n$, you want to show that $\lim_{n \to \infty} n^{1/n} = 1 $.
Here's a bit of magic:
By Bernoulli's inequality, $(1+\frac1{\sqrt{n}})^n \ge 1+n\frac1{\sqrt{n}} \ge 1+\sqrt{n} \gt \sqrt{n} = n^{1/2} $.
Raising both sides to the $2/n$ power, $((1+\frac1{\sqrt{n}})^n)^{2/n} > (n^{1/2})^{2/n} $ or $(1+\frac1{\sqrt{n}})^2 > n^{1/n} $ or $n^{1/n} < (1+\frac1{\sqrt{n}})^2 =1+2\frac1{\sqrt{n}}+\frac1{n} <1+3\frac1{\sqrt{n}} $.
Since $n^{1/n} > 1$ and $\lim_{n \to \infty}\frac1{\sqrt{n}} =0 $, $\lim_{n \to \infty} n^{1/n} = 1 $.
As to avoiding the use of L'Hospital's rule:
$$\ln(x)=O(\sqrt x)$$
$$\implies\lim_{x\to\infty}\frac{\ln(x)}x=\lim_{x\to\infty}\frac{\sqrt x}x$$
$$=\lim_{x\to\infty}\frac1{\sqrt x}=0$$
And so we have $\lim x^{1/x}=e^{\ln(x)/x}=e^0=1$
Let $x=1+\frac {1}{y}$ Where $y \rightarrow 0$
Then,
$\lim _{x\to \infty }x^{\frac{1}{x}} =\lim _{y\to 0 }{(1+\frac {1}{y})}^{\frac {y}{1+y}} = \lim _{y\to 0 }{(1+\frac {1}{y})}^{y\frac {1}{1+y}}$
$=\lim _{y \to 0}{1^{\frac {1}{1+y}}}=1$
