Basic evaluation of a limit 2 e

Question

How would evaluate the the following limit:

$$\lim_{n\to\infty} \frac {1}{n^{\frac{2}{n}}}$$

Would $n^{\frac{2}{n}} = n^0$ If this can not be evaluated like this, could I use l'hopitals rule, and if so how?

Answer 1

Hint:

$$1 \leqslant n^{1/n} = \exp\left( \frac{\log n}{n}\right) = \exp\left( \frac{2\log \sqrt{n}}{n}\right) \leqslant \exp\left( \frac{2\sqrt{n}}{n}\right) = \exp\left( \frac{2}{\sqrt{n}}\right) $$

Answer 2

Similar to RRL's:

By Bernoulli's inequality, $(1+\frac1{n^{1/2}})^n \ge 1+\frac{n}{n^{1/2}} \gt n^{1/2} $.

Raising to the $2/n$ power, $n^{1/n} \le (1+\frac1{n^{1/2}})^2 =1+\frac{2}{n^{1/2}}+\frac1{n} <1+\frac{3}{n^{1/2}} $.

Therefore $n^{2/n} <(1+\frac{3}{n^{1/2}})^2 =1+\frac{3}{n^{1/2}}+\frac{9}{n} =1+\frac{12}{n^{1/2}} $.

Note: It is possible to show by similar elementary methods that, for any integer $k > 1$, there is a $c(k)$ such that $n^{1/n} < 1+\frac{c(k)}{n^{1-1/k}} $ for large enough $n$.

Answer 3

I will show that $\lim\limits_{n\to\infty}n^\frac{1}{n}=1$. Then from the continuity of the function $f(x)=\frac{1}{x^2}$ at $x=1$, it will follow that The limit in question is equal to $1$.

Let us write $x_n=n^\frac{1}{n}-1$. Note that $x_n\geq0$ for all $n\geq1$. Then we have $$n=(x_n+1)^n=\sum_{k=0}^n\binom{n}{k}x_n^k\geq1+\frac{n(n-1)}{2}x_n^2,$$ so that $$0\leq x_n\leq\sqrt{\frac{2}{n}}\to0$$ as $n\to\infty$. Thus $0=\lim\limits_{n\to\infty}x_n=\lim\limits_{n\to\infty}\left(n^\frac{1}{n}-1\right)$ and therefore $\lim\limits_{n\to\infty}n^\frac{1}{n}=1$.