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How can I evaluate: $\lim_{h \to 0} \int_{-1}^{1}\frac{h}{h^2+x^2}~dx$

How I proceed: $\lim_{h \to 0} \int_{-1}^{1}\frac{h}{h^2+x^2}~dx=2\lim_{h \to 0} \frac{1}{h}\int_{0}^{1}\frac{1}{1+(\frac{x}{h})^2}~dx=2\lim_{h \to 0}\frac{1}{h}\arctan\frac{1}{h}$ Then how can I prooceed. Please help. Thank in advance.

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    Last expression will be $2\lim_{h \to 0}\arctan\frac{1}{h}$.I was wrong.2012-12-21

2 Answers 2

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Hint: $\lim_{x\to +\infty}\arctan x=\frac{\pi}2$ while $\lim_{x\to -\infty}\arctan x=-\frac{\pi}2$ Therefore, $\lim_{h \to 0^+}\frac1h\arctan\frac{1}{h}=\lim_{y \to +\infty}y\arctan y=(+\infty)\frac\pi 2=+\infty$ while $\lim_{h \to 0^-}\frac1h\arctan\frac{1}{h}=\lim_{y \to -\infty}y\arctan y=(-\infty)\frac{-\pi} 2=+\infty$ The required limit is $\lim_{h \to 0}\frac1h\arctan\frac{1}{h}=+\infty$

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    No.It's enough.2012-12-22
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$\lim_{h \to 0} \int_{-1}^{1}\frac{h}{h^2+x^2}~dx=2\lim_{h \to 0} \frac{1}{h}\int_{0}^{1}\frac{1}{1+(\frac{x}{h})^2}~dx=2\lim_{h \to 0} \int_{0}^{1/h}\frac{1}{1+(\frac{x}{h})^2}~d\left({x\over h}\right) $ $ 2\lim_{x\rightarrow 0} \arctan(1/x) = \pi \text{ as stated above.}$

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    Why change the upper limit in the integral? There's no substitution there!2012-12-21