# Evaluate: $\lim_{h \to 0} \int_{-1}^{1}\frac{h}{h^2+x^2}~dx$

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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Why can't you go on by yourself :-)? You have almost finished: there is no indeterminate form. –  Romeo Dec 21 '12 at 11:27
you put extra ${1 \over h}$ in the last –  Santosh Linkha Dec 21 '12 at 11:34
@experimentX: I forgot that in previous expresion –  Argha Dec 21 '12 at 11:36
Last expression will be $2\lim_{h \to 0}\arctan\frac{1}{h}$.I was wrong. –  Argha Dec 21 '12 at 12:05

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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If I take $y=\frac{1}{h}$ then I get $\lim_{y \to \infty}y \arctan y$.What I can do with this y. –  Argha Dec 21 '12 at 11:33
@Argha As you said, you are looking for the limit of $\arctan\frac1h$ and not $\frac1h\arctan\frac1h$ –  Nameless Dec 21 '12 at 11:43
No,I want $\frac{1}{h}\arctan \frac{1}{h}$ not $\arctan \frac{1}{h}$ –  Argha Dec 21 '12 at 11:47
So, I conclude limit exists and equal to $+\infty$ –  Argha Dec 21 '12 at 11:51
now I understand the problem. Thank you –  Argha Dec 21 '12 at 11:54
$$\lim_{h \to 0} \int_{-1}^{1}\frac{h}{h^2+x^2}~dx=2\lim_{h \to 0} \frac{1}{h}\int_{0}^{1}\frac{1}{1+(\frac{x}{h})^2}~dx=2\lim_{h \to 0} \int_{0}^{1/h}\frac{1}{1+(\frac{x}{h})^2}~d\left({x\over h}\right)$$ $$2\lim_{x\rightarrow 0} \arctan(1/x) = \pi \text{ as stated above.}$$
why don't you change limit of integration $(0,1)$ to $(0,1/h)$ –  Argha Dec 21 '12 at 11:44
$\lim_{h \to 0} \int_{0}^{1/h}\frac{1}{1+(\frac{x}{h})^2}~d\left({x\over h}\right) = \lim_{h \to 0} \arctan (1/h) - \arctan 0$ which is same –  Santosh Linkha Dec 21 '12 at 11:59
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