Calculating limit

Question

I need confirmation for the following question.

Which of the following statements is true?

$1.$ $\lim_{x\to\infty}\frac{\log x}{x^{1/2}}=0$ and $\lim_{x\to\infty}\frac{\log x}x=\infty$

$2.$ $\lim_{x\to\infty}\frac{\log x}{x^{1/2}}=\infty$ and $\lim_{x\to\infty}\frac{\log x}x=0$

$3.$ $\lim_{x\to\infty}\frac{\log x}{x^{1/2}}=0$ and $\lim_{x\to\infty}\frac{\log x}x=0$

$4.$ $\lim_{x\to\infty}\frac{\log x}{x^{1/2}}=0$ but $\lim_{x\to\infty}\frac{\log x}x$ does not exist.

So, I am thinking that however large $x$ be, $\log x$ will remain small. So, a small quantity over a very large one would give us $0$ as result. And sqrt of a large number is also a large number. So, answer should be $3).$

Am I correct here?

Well, except that $\log x$ will not remain small. It grows much slower than any power though. — Daniel Fischer, Aug 16 '13 at 18:32

lab bhattacharjee · Answer 1 · 2013-08-16T18:44:34.510

2

HINT:

As $\displaystyle\lim_{x\to\infty^+}\frac{\ln x}{x^n}$ is of the form $\frac\infty\infty$ for $n>0$

Applying L'Hospital's Rule

$\displaystyle\lim_{x\to\infty^+}\frac{\ln x}{x^n}=\lim_{x\to\infty^+}\frac1x \cdot\frac1{nx^{n-1}}=\frac1n\cdot\frac1{\lim_{x\to\infty^+}x^n}=0$ if $n>0$

edited Aug 16 '13 at 18:44

answered Aug 16 '13 at 18:32

lab bhattacharjee

274,582

log is not defined for -ve values, so it must be $\infty^+$. But why specifically for non-integers? I would assume it should be so for every n. – aarbee Aug 16 '13 at 18:43
@Ramit, thanks for the observation. Rectified – lab bhattacharjee Aug 16 '13 at 18:45

score 2 · Accepted Answer · answered Aug 16 '13 at 18:33

Your choice is correct: $\log x$ grows more slowly than any positive power of $x$. It’s not true that $\log x$ remains small, though. You can make it as large as you want: assuming that this is the natural log, $\log e^x=x$. The $0$ limits are easily verified using l’Hospital’s rule, if you’re familiar with it.

Calculating limit

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