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I was trying to solve a question that was uploaded here (Computing $\lim_{n\rightarrow \infty} \frac{1+\sqrt{2}+\sqrt[3]{3}+\cdots+\sqrt[n]{n}}{n}$.):

$$\lim_{x \to \infty} \frac {1 + 2^{1/2} + 3^{1/3} +...+ x^{1/x}}{x}$$

So i used the identity $$\lim_{x \to \infty} x^{1/x} = 1$$ and tried to prove it myself.

Since $$(\sqrt[x]{x})^x = x\implies f(x) = \sqrt[x]{x} = \frac x{(\sqrt[x]{x})^{x-1}}$$

I tried to prove that there is a horizontal asymptote at $y=1$ when $\lim\limits_{x \to \infty} x$ using the latter function without using L'Hôpital's rule. I want to emphasize that I am not currently trying to solve the first limit, but trying to prove the asymptote of the function $f(x)$.

Any suggestions?

Omer Paz
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    What is the question? If it is in your title then this is a duplicate. – Jyrki Lahtonen Apr 20 '18 at 20:06
  • Your idea seems to be a good approach, namely that once you know $\lim_{n\to \infty} n^{1/n} = 1$, the more elaborate limit treated in the linked Question becomes a straightforward consequence. Perhaps you are having difficulty carrying through this approach? – hardmath Apr 20 '18 at 20:11

2 Answers2

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Note that

$$x^\frac1x=e^{\frac{\log x}{x}}\to 1$$

indeed by standard limit

$$\frac{\log x}{x}\to 0$$

To prove the latter set $x=e^y$ with $y\to \infty$ then

$$\frac{\log x}{x}=\frac{\log e^y}{e^y}=\frac{y}{e^y}\to 0$$

which is true since eventually $e^y\ge y^2$ (can be proved by induction) and then

$$0\le \frac{y}{e^y}\le \frac{y}{y^2}\le \frac1y \to 0$$

user
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You may rty ab oveo if you like:

For any $\epsilon>0$ and $n\in\Bbb N$, we have $(1+\epsilon)^n\ge 1 + n\epsilon+\frac{n(n-1)}{2}\epsilon^2$. This is $>n+1 $ as soon as $n>\frac2{\epsilon^2}+1$. Hence $1<(n+1)^{1/n}<1+\epsilon$ for all $n>\frac2{\epsilon^2}+1$. Then for all $x>\frac2{\epsilon^2}+2$, we find $ n\in \Bbb N$ with $n\le x<n+1$ and $n>\frac2{\epsilon^2}+1$. We conclude that for such $x$, $$ 1\le x^{1/x}\le(n+1)^{1/n}<1+\epsilon.$$ Therefore, $$ \lim_{x\to\infty}x^{1/x}=1.$$

Hagen von Eitzen
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