I encountered a problem: calculate $\lim\limits_{x\to\infty} x^{1/x}$
My solution is quite weird and I find no document explaining whether I am right or wrong!
My solution:
As $\lim\limits_{x\to\infty} 1/x = 0$
then $\lim\limits_{x\to\infty} x^{1/x} = \lim\limits_{x\to\infty} x^0 = 1$
My question is: Whether I am right or wrong, and could you please explain why?
You need to be careful; it looks like you treated the $x$ in the base as a constant with respect to the limit, while it is not. The typical way to evaluate the above is apply the continuous transform $\log$ to deal with the exponent $$ L=\lim_{x\rightarrow \infty}x^{1/x}\Rightarrow \log L=\lim_{x\rightarrow \infty}\frac{1}{x}\log x=\lim_{x\rightarrow \infty}1/x=0\Rightarrow L=e^0=1$$ by L'Hôpital
$$\lim_{x \to \infty} x^{\frac 1x} = \lim_{x \to \infty} \exp{\frac{\log x}{x}}$$
Put the limit inside the $\exp$ (by continuity) and let $x = e^u$.
Then for large enough $x$ $$0 < \frac{\log x}{x} = \frac{u}{e^u} \leq \frac{u}{1 + u + \frac{u^2}{2}} \xrightarrow[u \to \infty \Longleftrightarrow x \to \infty]\ 0 $$
implies the limit is equal to $1$.
Using l'Hôpital's rule:
$$\lim_{x\to\infty}x^{\frac{1}{x}}=\exp\left[\lim_{x\to\infty}\frac{\ln(x)}{x}\right]=\exp\left[\lim_{x\to\infty}\frac{\frac{1}{x}}{1}\right]=\exp\left[\lim_{x\to\infty}\frac{1}{x}\right]=\exp\left[0\right]=e^0=1$$
Let $y = x^{1/x}$.
Then $\ln(y) = \dfrac{1}{x}\ln(x)$, implying $$\lim_{x \to \infty}\ln(y)=\lim_{x \to \infty}\dfrac{\ln(x)}{x}\text{.}$$ By continuity of $\ln$, we have $$\lim_{x \to \infty}\ln(y) = \ln\left( \lim_{x \to \infty}y\right)$$ so that $$\lim_{x \to \infty}y = e^{\lim_{x \to \infty}\frac{\ln(x)}{x}}\text{.}$$ Now by L-Hospital, $$\lim_{x \to \infty}\dfrac{\ln(x)}{x} \overset{H}{=}\lim_{x \to \infty}\dfrac{1/x}{1} = 0$$ thus giving $$\lim_{x \to \infty}y = e^0 = 1\text{.}$$
