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 $1\over x^{1/2}$.

$3$ is correct as $\log x$ grows slower than any $x^n$. So $x^{-1}$ and $x^{-\frac{1}{2}}$ will manage to pull it down to $0$. And $\displaystyle\lim_{x\to\infty}{1\over x}=\lim_{x\to\infty}{1\over \sqrt{x}}=0$.
