Using the logarithm, find $\lim_{n\rightarrow \infty} n^{1/n}$

Question

Using the logarithm find $$\lim_{n\rightarrow \infty} n^{1/n}$$

Here is my attempt: \begin{align} \lim_{n\rightarrow \infty} n^{1/n} &= \lim_{n\rightarrow \infty} \exp(\ln(n^{1/n})) \\ &= \lim_{n\rightarrow \infty} \exp\Bigl(\frac{\ln(n)}{n}\Bigr) \\ &= \exp \Bigl(\lim_{n\rightarrow \infty} \frac{\ln(n)}{n}\Bigr) \end{align}

Not sure if I can do this last step. And even if I can, not sure how to go further from here.

Answer 1

The last step is proper because exponentiation is continuous. Use L'Hopital's Rule to evaluate the last limit.

Answer 2

Without L'Hospital's rule:

Prove that, for $x>1$, $\ln x<2\sqrt x$. Deduce the limit of $\frac{\ln x}x$ as $x$ tends to $+\infty$.

Answer 3

Using essentially nothing: Let $n^{1/n} = 1 +a_n.$ Think of $n$ large and apply the binomial theorem: $n = (1+a_n)^n = 1+na_n + [n(n-1)/2]a_n^2 + \dots$ Thus $n \ge [n(n-1)/2]a_n^2,$ which implies $1/[(n-1)/2] \ge a_n^2.$ This implies $a_n\to 0,$ hence $n^{1/n}\to 1.$