Check if $\lim_{x\to\infty}{\log x\over x^{1/2}}=\infty$, $\lim_{x\to\infty}{\log x\over x}=0$ Could anyone tell me which of the following is/are true?

  
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*$\lim_{x\to\infty}{\log x\over x^{1/2}}=0$, $\lim_{x\to\infty}{\log x\over x}=\infty$
  
*$\lim_{x\to\infty}{\log x\over x^{1/2}}=\infty$, $\lim_{x\to\infty}{\log x\over x}=0$
  
*$\lim_{x\to\infty}{\log x\over x^{1/2}}=0$, $\lim_{x\to\infty}{\log x\over x}=0$
  
*$\lim_{x\to\infty}{\log x\over x^{1/2}}=0$, but $\lim_{x\to\infty}{\log x\over x}$ does not exists.

For I know $\lim_{x\to\infty}{1\over x}$  exist and so is for $1\over x^{1/2}$.

In order to generalize the function, I'll write it as:
$$\lim_{x\to\infty}\frac{\ln(x)}{x^n}$$
For $n>0$. Using L'Hôpital's rule gives:
$$\lim_{x\to\infty}\frac{\ln(x)}{x^n}\\=\lim_{x\to\infty}\frac{\frac{1}{x}}{nx^{n-1}}\\=\frac{1}{n}\lim_{x\to\infty}\frac{1}{x^n}\\=0$$
