Limit of Integral Evaluation: Tips and Tricks

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How would I go about evaluating something like this?

<br /> \[<br /> \lim_{n\to+\infty} n \int_0^{+\infty} \dfrac{\sin\left(\dfrac{x}{n}\right)}{x(1+x^2)}\, dx<br /> \]<br />
 
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Let I(n) be the definite integral.

Rewrite your expression as :
\frac{I(n)}{\frac{1}{n}}

Note that you can use L'hopitals rule here, and that you may interchange the operation of differentiation with respect to "n" and integration with respect to "x".
 
Another way might be to write:
\sin(\frac{x}{n})=\frac{x}{n}++++

Inserting, and simplifying, the limit will be the same as the above.
 
\frac{ n \sin\left(\frac{x}{n}\right)}{x(1+x^2)} \rightarrow \frac{1}{1+x^2} uniformly, so we can bring everything inside the integral.
 
Thanks everyone!
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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