lim-how to write it down correct?

Question

I want to show that

$\lim_{n\to\infty} n^{1/n}=1$

I know why it is $1$. Cause $\frac{1}{n}\to 0$ But how can I write that correct?

Answer 1

For large enough $n,$

$$0 <\ln (n)<\sqrt {n} $$

$$0 <\ln (n^\frac 1n)<\frac {1}{\sqrt {n}} $$

thus $$e^0 <n^\frac 1n <e^\frac {1}{\sqrt {n}} $$

and squeeze.

Answer 2

$$n^{1/n}=e^{ln(n)/n} \underset{n \rightarrow +\infty}{\rightarrow}e^{0}=1$$

Answer 3

It is not enough to notice that the exponent is going to zero. If the base of the exponentiation is going to infinity, the limit may fail to be one. For example, $(n^n)^{1/n}$ goes to infinity, even though the exponent goes to zero.

In your case, however, you are right, the limit is one. To compute it, use l'Hôpital's on the exponential indeterminate form, by first taking the logarithm.

In detail, if $L=\lim_{n\to\infty}n^{1/n}$, then $$\log L = \log(\lim_{n\to\infty}n^{1/n})=\lim_{n\to\infty}\log(n^{1/n})=\lim_{n\to\infty}\frac{\log(n)}{n},$$ the latter fraction is seen to be an $\infty/\infty$ indeterminate form, which meets the criteria for l'Hôpital's method. So then taking derivatives

$$\log L = \lim_{n\to\infty}\frac{\log(n)}{n}=\lim_{n\to\infty}\frac{1/n}{1}\to 0.$$

Thus we have that $\log L=0,$ or $L=1.$

Answer 4

Prove that $$1<\sqrt[n]{n}<1+\sqrt{\frac{2}{n}}$$ and use the sandwich.