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

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 ?

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  • $\begingroup$ Fundamental theorem of calculus is the only thing you need here. $\endgroup$ – Claude Leibovici Apr 17 '14 at 13:13
  • $\begingroup$ @Claude Leibovici So the function is strictly increasing, but the rate of increase changes. $\endgroup$ – OBDA Apr 17 '14 at 13:16
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does the integral $\displaystyle\int_1^x\ln(1+\frac{1}{\sqrt{t}})dt$ really diverge ?

It surely does: $$t\geqslant1\implies\log\left(1+\frac1{\sqrt{t}}\right)\geqslant\frac{\log2}{\sqrt{t}}$$

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  • $\begingroup$ Do you use Taylor ? $\endgroup$ – OBDA Apr 17 '14 at 13:25
  • $\begingroup$ Taylor works. Convexity might be simpler. $\endgroup$ – Did Apr 17 '14 at 16:47

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