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I'm having trouble with this problem - I don't even know how to begin. Thoughts? Solutions with explanation? Please help!

Let $f$ be a bounded continuous function on $\mathbb{R}$. Prove that

$$ \lim_{n \to \infty} \frac{n}{\pi} \int_{\mathbb{R}} \frac{f(t)}{1+n^2t^2}dt = f(0).$$

Dmitri Valentine
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    this question is answered here http://math.stackexchange.com/questions/997709/integral-of-bounded-continuous-function-on-r/997741#997741 – David Holden Oct 31 '14 at 11:30
  • Thanks for your response, though I have a question.

    The integral I'm asking for is slightly different. It's being multiplied by n, which is approaching infinity. Does that affect your argument?

     – Dmitri Valentine Oct 31 '14 at 11:42
  • the $n$ is necessary because you want $\frac{n}{\pi} \int_{\mathbb{R}} \frac{dt}{1+n^2t^2}dt = 1$ if you make the substitution $x=nt$ in $\frac{n}{\pi} \int_{\mathbb{R}} \frac{f(t)}{1+n^2t^2}dt $ you will see that the $n$ cancels and shows up only as a scaling factor in $\frac1{\pi} \int_{\mathbb{R}} \frac{f(\frac{x}{n})}{1+x^2}dx $ – David Holden Oct 31 '14 at 12:00

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