# Prove limit of integral

1. Mar 11, 2008

### sinClair

Never mind, got it.

Last edited: Mar 11, 2008
2. Mar 11, 2008

### MasterTinker

I can possibly help you by generating a convenient little case for you:

Consider f(x) = c where c is a constant.

Take the integral and look at your limit, you should have:

$$\mathop{\lim}\limits_ {n \to \infty}n^\alpha\int_{0}^{1/n^\beta}cdx$$

= $$\mathop{\lim}\limits_ {n \to \infty}cn^\alpha\c/n^\beta$$

= $$\mathop{\lim}\limits_ {n \to \infty}c/n^{\beta-\alpha}$$

And given $$\alpha<\beta$$, what do you know about this limit? (Notice that this still works for $$\alpha<0$$, which you need to consider given the restrictions on $$\alpha$$).

I think you can eventually generalize from there. Happy integrating :tongue:

3. Mar 11, 2008

### sinClair

Thanks for the suggestion Tinker. Yes that is a convenient case but that also involves using the fundamental theorem of calculus to actually integrate. But for an arbitrary function it's impossible to explicitly calculate the integral like that and get a nice expression to take the limit. So I'm actually thinking of doing it using the definition of Riemann sum.