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I have to solve a limit of a definite integral, but honestly I have absolutely no idea how to do it. I solved this integral (I just solved $\int \frac{1}{t}$, $-\int \ln(1+t)$, and $\int \ln(t)$) but the result is so big it's not even possible to solve this limit with 'brute force' method. I'm pretty sure there is a theorem or something which helps solving it immensely, but I'm not familiar with it... Can you help me solve it?

$$\lim_{x \to \infty} \frac1{x^2} \int_{1}^{x^2} \frac{2}{t} - \ln\left(\frac{1+t}{t}\right)\ \mathrm dt$$

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    $\begingroup$ Did you integrate the top? $\endgroup$ – mysatellite Jun 22 '15 at 23:21
  • $\begingroup$ The first step would be to make it readable so everyone can read your limit without a magnifier :P $\endgroup$ – AlexR Jun 23 '15 at 10:43
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Hint: L'Hopital + FTC makes short work of this, especially remembering we only need the denominator $\to \infty$ to use it.

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for $b$ large, i want to see if we can evaluate $$\begin{align}\int_{1}^{b}\left( \frac{2}{t} - \ln\left(\frac{1+t}{t}\right)\right)\, dt &= \left(2\ln t - (1+t)\ln(1+t) +(1+t)+t\ln t - t\right)_1^{b}\\ &= 2\ln b - (1+b)\ln(1+b)+b\ln b+2\ln 2\\ &=2\ln b-(1+b)\left(\ln b + \frac1b -\frac1{2b^2} + \cdots\right)+b\ln b+2\ln 2\\ &=2\ln b-b\left(\ln b + \frac1b -\frac1{2b^2} + \cdots\right)\\ &-\left(\ln b + \frac1b -\frac1{2b^2} + \cdots\right)+b\ln b+2\ln 2\\ &=\ln b+2\ln 2-1-\frac1{2b}+\cdots \end{align}$$

therefore $$\frac1b \int_{1}^{b}\left( \frac{2}{t} - \ln\left(\frac{1+t}{t}\right)\right)\, dt =\frac{\ln b + 2\ln 2 - 1+\cdots}{b} \to 0 \text{ as } b \to \infty. $$

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