Sequence involving $\sqrt[n]{n}$ and $\text{log}$

Question

I know that the sequence $\sqrt[n]{n}$ converges to 1 and that $\text{log}(\sqrt[n]{n})$ thus converges to 0 as $n\to\infty$ since the logarithmic function is continuous. But how can I calculate the limits as $n\to\infty$ of the following sequences?

(a) $\frac{\sqrt[n]{n^2}-1}{\text{log}(\sqrt[n]{n})}$

(b) $\frac{\sqrt[n]{1/n}-1}{\text{log}(\sqrt[n]{n})}$

Any help would be much appreciated!

Answer 1

(a) Write this quotient as

$$\frac{e^{(1/n)\ln n^2}-1}{(1/n)\ln n} = \frac{e^{2(1/n)\ln n}-1}{(1/n)\ln n}.$$

As $n\to \infty,$ $(1/n)\ln n\to 0.$ So the limit of the above has the form

$$\lim_{u\to 0}\frac{e^{2u}-1}{u}.$$

That should look familiar.

Answer 2

For any continuous function $f(x)$ such that $f(x)\rightarrow 1$ as $x\rightarrow a$ we can say that $\lim_{x\rightarrow a} \frac{\ln (f(x))}{f(x)-1}=1$, i.e., $\ln(f(x))$ starts approaching $f(x)-1$. Make the same replacements in your question as

(i) $\frac{n^{2/n}-1}{\ln{n^{1/n}}}\rightarrow\frac{n^{2/n}-1}{n^{1/n}-1}\rightarrow n^{1/n}+1\rightarrow 2$ as $n\rightarrow \infty$.

(ii) Using the above replacement for $\ln$ we can write it as $\frac{1-n^{1/n}}{n^{1/n}(n^{1/n}-1)}\rightarrow \frac{-1}{n^{1/n}}\rightarrow -1$ as $ n\rightarrow\infty$.