Limit of the root $\sqrt[k]{k}$

Question

Haow can I calculate this limit? $$L=\lim_{k\to\infty}\sqrt[k]{k}$$ I suppose its value is one, but how can I prove it? Thanks

Answer 1

HINT: Note that $$\ln L=\lim_{k\to \infty}\frac{\ln k}{k}$$ Now use L'Hospital's rule.

Answer 2

Let $x_n=n^{1/n}=1+\alpha_n$

It can easily be proved that $x_n\ge 1$ so $\alpha_n\ge 0$

Then we have,

$n=(1+\alpha_n)^n=\sum_{i=0}^{\infty }\binom{n}{i}\alpha_n^i\ge 1+n\alpha_n+\frac{n(n-1)}{2}\alpha^n\ge \frac{n(n-1)}{2}\alpha_n^2$

So we have,

$n\ge \frac{n(n-1)}{2}\alpha_n^2\Rightarrow \alpha_n^2\le \frac{2}{n-1}$

Now we have $0\le \alpha_n^2\le \frac{2}{n-1}$ and this implies $\lim_{n\to \infty }\alpha_n=0\Rightarrow \lim_{n\to \infty }x_n=1$(By squeeze theorem)

Answer 3

Continuing from pritam's answer:

$\ln L=\lim_{k\to \infty}\frac{\ln k}{k}=$(Via L'Hospital's Rule: take derivative of numerator and denominator)

$\implies \ln L=\lim_{k\to \infty}\frac{\frac{1}{k}}{1}=\implies \ln L=\lim_{k\to \infty}\frac{1}{k}\implies\ln L=0\implies L=e^0=1$