Brodie Beck

2022-09-05

How do I calculate the limit of this log function?
I've been stuck with calculating the limit of the following problem for a while now. Can you help?
$\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{\mathrm{log}\left(x\right)+1}}{\mathrm{log}\left(\mathrm{log}\left(x\right)\right)}=$

Substitute $u=\mathrm{log}\left(x\right)$. Then, since you get a $\frac{\mathrm{\infty }}{\mathrm{\infty }}$ expression, you can apply L'Hospital's rule:
$\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{\mathrm{log}\left(x\right)+1}}{\mathrm{log}\left(\mathrm{log}\left(x\right)\right)}=\underset{u\to \mathrm{\infty }}{lim}\frac{\sqrt{u+1}}{\mathrm{log}\left(u\right)}=\underset{u\to \mathrm{\infty }}{lim}\frac{u}{2\sqrt{u+1}}=\mathrm{\infty }$

reemisorgc

$\underset{x\to \mathrm{\infty }}{lim}\frac{\sqrt{\mathrm{log}x+1}}{\mathrm{log}\left(\mathrm{log}x\right)}=\underset{x\to \mathrm{\infty }}{lim}\frac{\frac{1}{2x\sqrt{\mathrm{log}x+1}}}{\frac{1}{x\mathrm{log}x}}=\underset{x\to \mathrm{\infty }}{lim}\frac{\mathrm{log}x}{2\sqrt{\mathrm{log}x+1}}=\mathrm{\infty }$