How to calculate the limit $\lim_{x\to\infty} (x^{1/n}-\ln(x))$

Question

As in the title, I am having trouble with this seemingly easy limit. I cannot seem to transform it into a form that would let us use L'Hôpital. I also cannot see anything I can multiply by to simplify the limit. I was wondering if we can perhaps use the fact that $$ \lim_{x\to\infty}\frac{\ln(x)}{x^{\frac{1}{n}}}=0. $$ Or maybe raise the limit to the $n$th power but that would require a messy binomial. I am probably missing something obvious but I would appreciate any hints.

Answer 1

Yes, you can use that fact: $$ \lim_{x\to\infty}(x^{1/n}-\ln x)= \lim_{x\to\infty}\left(\frac{x^{1/n}}{\ln x}-1\right)\ln x $$ The limit of the part in parentheses is $\infty$.

Answer 2

We have

$$x^{1/n}-\ln(x)=\ln x\left(\frac{x^{1/n}}{\ln x}-1\right)\to \infty$$

indeed $\ln x \to \infty$ and for $x=e^y$ with $y \to \infty$

$$\frac{x^{1/n}}{\ln x}=\frac{e^{y/n}}{y}=\frac1n\frac{e^{y/n}}{y/n}\to \infty$$

indeed eventually as $t \to \infty$ we have $e^t>t^2$ and then

$$\frac{e^{y/n}}{y/n}\ge \frac{(y/n)^2}{y/n}=\frac y n \to \infty$$

Answer 3

Hint:

Set $t=x^{\tfrac1n}$, so the expression can be re-written as $$x^{\tfrac1n}-\ln x=t-n\ln t=t\Bigl(1-n\,\frac{\ln t}t\Bigr),$$ and use that near $\infty$, $\:\ln t=o(t)$.

Answer 4

Use that: $$\lim_{n\to k} {[f(n)-g(n)]}=\lim_{n\to k}{[f(n)]}-\lim_{n\to k}{[g(n)]}$$