Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
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
up vote 2 down vote accepted

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 $$\lim_{h \to 0}\frac1h\arctan\frac{1}{h}=+\infty$$

share|cite|improve this answer
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

$$\lim_{h \to 0} \int_{-1}^{1}\frac{h}{h^2+x^2}~dx=2\lim_{h \to 0} \frac{1}{h}\int_{0}^{1}\frac{1}{1+(\frac{x}{h})^2}~dx=2\lim_{h \to 0} \int_{0}^{1/h}\frac{1}{1+(\frac{x}{h})^2}~d\left({x\over h}\right) $$ $$ 2\lim_{x\rightarrow 0} \arctan(1/x) = \pi \text{ as stated above.}$$

share|cite|improve this answer
why don't you change limit of integration $(0,1)$ to $(0,1/h)$ – Argha Dec 21 '12 at 11:44
yeah!! sorry forgot about that!! – Santosh Linkha Dec 21 '12 at 11:48
$\lim_{h \to 0} \int_{0}^{1/h}\frac{1}{1+(\frac{x}{h})^2}~d\left({x\over h}\right) = \lim_{h \to 0} \arctan (1/h) - \arctan 0$ which is same – Santosh Linkha Dec 21 '12 at 11:59
this seems to work – Santosh Linkha Dec 21 '12 at 12:01
I delete my comment.You are right. – Argha Dec 21 '12 at 12:01

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.