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I tried to find the limit of this function, without any success,

$$\lim _{x\to \infty }\frac{\displaystyle \left(\int _{10}^x\frac{e^{\sqrt{t}}}{t}dt\right)^2}{e^x}$$

I tried to solve it by L'Hôpital's rule $\left(\frac{\infty }{\infty }\right)$ but there is no way to do it.

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    $\begingroup$ What's squared? The integrand or the integral? $\endgroup$ – user228113 Jun 28 '15 at 9:53
  • $\begingroup$ The integral, edited, sorry. $\endgroup$ – Matan Jun 28 '15 at 9:56
  • $\begingroup$ Now what's squared? $\endgroup$ – math110 Jun 28 '15 at 10:22
  • $\begingroup$ @math110 edited again. $\endgroup$ – Matan Jun 28 '15 at 10:23
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You can use L'Hospital's rule as follows: $$\lim _{x\to \infty }\frac{\displaystyle \int _{10}^x\frac{e^{\sqrt{t}}}{t}dt}{e^{x/2}}=\lim_{x\to\infty}\frac{\displaystyle\frac{e^\sqrt{x}}{x}}{\frac12e^{x/2}}=0.$$ The first equality is due to the fundamental theorem of calculus: $$\frac{d}{dx}\int_{x_0}^xf(t)dt=f(x),\text{when }f\text{ is continuous.}$$ The second one is due to the fact that $xe^{x/2}$ goes much faster than $e^\sqrt x$.

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  • $\begingroup$ Thanks but what about the squared? can i ignore it? $\endgroup$ – Matan Jun 28 '15 at 10:34
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    $\begingroup$ @Matan $0^2 = 0$ $\endgroup$ – Eclipse Sun Jun 28 '15 at 10:44
  • $\begingroup$ @EclipseSun,you answer is more simple+1 $\endgroup$ – math110 Jun 28 '15 at 10:44
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    $\begingroup$ @EclipseSun. May I suggest you add to your nice answer that the FTC is used. I suppose it could help th OP. Thanks. $\endgroup$ – Claude Leibovici Jun 28 '15 at 11:21
  • $\begingroup$ @ClaudeLeibovici ok $\endgroup$ – Eclipse Sun Jun 28 '15 at 14:52
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Hint $$e^{\sqrt{t}}\le e^{\sqrt{x}},t<x$$ then we have $$\left(\int_{10}^{x}\dfrac{e^{\sqrt{t}}}{t}dt\right)^2\le e^{2\sqrt{x}}\cdot\left(\int_{10}^{x}\dfrac{1}{t}dt\right)^2=e^{2\sqrt{x}}(\ln{x}-\ln{10})^2$$ and Note this limits $$\lim_{x\to+\infty}\dfrac{\ln^2{x}}{e^{x-2\sqrt{x}}}=0$$ then you limits is equal $0$

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  • $\begingroup$ I don't understand, what did you do with the Integral? thanks $\endgroup$ – Matan Jun 28 '15 at 10:27
  • $\begingroup$ Now I have add some hints $\endgroup$ – math110 Jun 28 '15 at 10:29
  • $\begingroup$ nice! but is it legal to do such this thing?, you take out e from the integral. :( $\endgroup$ – Matan Jun 28 '15 at 10:32
  • $\begingroup$ It is clear legal $\endgroup$ – math110 Jun 28 '15 at 10:33
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    $\begingroup$ do you known this well known? if $f,g>0,x\in[a,b]$ and such $f(x)<M$,then have $\int_{a}^{b}f(x)g(x)dx\le M\int_{a}^{b}g(x)dx$$ $\endgroup$ – math110 Jun 28 '15 at 10:37

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