Having trouble solving the below limit. I tried L'Hôpital's Rule, but haven't gotten very far. Any ideas? $$\lim _{x\to \infty }\left(\frac{2^{\frac{1}{10}\left(log_2x\right)^{\frac{1}{2}}}}{log_2x}\right)$$

I made substitutions, $$\lim _{x\to \infty }\left(\frac{2^{\frac{\sqrt{x}\:}{10}}}{x}\right)$$ However, I am still stuck. Any help would be much appreciated!

  • 1
    $\begingroup$ Try another substitution after the first one, $x \to x^2$ and then L'Hôpital's rule twice. $\endgroup$
    – jjagmath
    Sep 18 at 10:58
  • $\begingroup$ Or continue on the line you've started, using $2^{\frac{\sqrt{x}}{10}=e^{\sqrt{x}\frac{\log 2}{10}}}\geqslant (\sqrt{x}\frac{\log 2}{10})^4/4!$ so that the thing of interest exceeds a constant multiple of $x$. $\endgroup$ Sep 18 at 11:02
  • $\begingroup$ The limit is of the form $\frac{a^t}{t^2}$ as $t \to \infty$, where $a>1$. $\endgroup$ Sep 18 at 11:20

1 Answer 1


$$\lim _{x\to \infty }\left(\frac{2^{\frac{1}{10}\left(log_2x\right)^{\frac{1}{2}}}}{log_2x}\right)$$ Apply substitution $x\mapsto x^4$ and use L'Hôpital's Rule: $$\lim _{x\to \infty }\frac{\frac{1}{10} 2^{\frac{x}{10}}\ln 2}{2x}$$ Especially after $x>10$, it is clear that the numerator is much greater than the denominator. Thus, the limit diverges to infinity. In fact, we can make the denominator look even more embarrassingly tiny by continuing to make substitutions: $$=\frac{\ln 2}{20}\lim _{x\to \infty }\frac{2^{\frac{x}{10}}}{x}$$ $$=\frac{\ln 2}{200}\lim _{x\to \infty }\frac{2^x}{x}$$

Thus, we have $$\lim _{x\to \infty }\left(\frac{2^{\frac{1}{10}\left(log_2x\right)^{\frac{1}{2}}}}{log_2x}\right)\color{red}{\text{ diverges to positive infinity}}$$


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