# Evaluate $\lim\limits_{x\to\infty}\frac{1}{\sqrt{x}}\int_1^x\ln(1+\frac{1}{\sqrt{t}})dt$

$\lim\limits_{x\to\infty}\frac{1}{\sqrt{x}}\displaystyle\int_1^x\ln(1+\frac{1}{\sqrt{t}})dt=?$

If the limit exists with l'Hopital i get

$\displaystyle\lim\limits_{x\to\infty}\frac{\ln(1+\frac{1}{\sqrt{x}})}{\frac{1}{2\sqrt{x}}}=2$

but why can I apply l'Hopital, I mean does the integral $\displaystyle\int_1^x\ln(1+\frac{1}{\sqrt{t}})dt$ really diverge ?

How can I show that ?

• Fundamental theorem of calculus is the only thing you need here. – Claude Leibovici Apr 17 '14 at 13:13
• @Claude Leibovici So the function is strictly increasing, but the rate of increase changes. – OBDA Apr 17 '14 at 13:16

does the integral $\displaystyle\int_1^x\ln(1+\frac{1}{\sqrt{t}})dt$ really diverge ?
It surely does: $$t\geqslant1\implies\log\left(1+\frac1{\sqrt{t}}\right)\geqslant\frac{\log2}{\sqrt{t}}$$