Finding $\lim_{n \rightarrow \infty} \frac{\log n}{\sqrt{n}}$

Question

Compute $$\lim_{n \rightarrow \infty} \frac{\log n}{\sqrt{n}}$$ It seems pretty obvious, but I have tried Stolz-Cesaro and other tricks and I still can't get a solution.

Answer 1

The fact you should remember is that $\log$ function will be bitten by any positive order of $x$, i.e., $x^\alpha$ for any $\alpha>0$. In another word, for any $\alpha>0$, $x^\alpha$ will growth faster then $\log x$ at infinity. Hence, the limit of you problem is $0$.

Answer 2

Why not use L'Hopital's rule?

\begin{equation} \lim_{n\rightarrow\infty}\text{log}(n)/n^{\frac{1}{2}}=\lim_{n\rightarrow\infty}\frac{\frac{1}{n}}{\frac{1}{2n^{1/2}}}=\lim_{n\rightarrow\infty}\frac{1}{2n^{1/2}}=0 \end{equation}

Answer 3

$$\lim_{x\to +\infty}\frac{\log x}{\sqrt{x}}=\lim_{t\to +\infty} t\, e^{-t/2} = 0.$$

Answer 4

$$\lim_{n\to\infty}{\log n\over\sqrt x}=\lim_{n\to\infty}{1/x\over \frac12x^{-1/2}}$$

by L'hopital $$\lim_{n\to\infty}{1/x\over \frac12x^{-1/2}}=\lim_{n\to\infty}\frac2{\sqrt x}=0.$$

Answer 5

Hint:

$$\log n^{1/4} < n^{1/4} \Rightarrow \log n < 4n^{1/4} \Rightarrow \frac{\log n}{\sqrt{n}} < \frac{4}{\sqrt[4]{n}} $$

Take $n \to \infty$.