# Quick series and integral Q

1. Feb 9, 2009

### insynC

Hi all. I've just come across two issues in some problems I have been doing and although I know the results, I can't remember the reason why or how to show it. Hoping someone can point me in the right direction.

1. The problem statement, all variables and given/known data

1: lim[n->∞] f(g(n)) = f(lim[n->∞]g(n))

2. Integral terminals [0,a]: ∫dx/(x(a-x))

3. The attempt at a solution

1: I'm pretty sure this is true given f is continuous. I'm sure there's some series theorem that establishes it but I haven't been able to find anything in my notes.

2: From memory I think this integral comes out to be something nice like pi or pi/2, but I couldn't figure it out with trig substitution.

Thanks

2. Feb 9, 2009

### HallsofIvy

Staff Emeritus
That's pretty much the definition of continuous!

What "trig substitution"? Do it by "partial fractions":
$$\frac{1}{x(a-x)}= \frac{1}{a}\left(\frac{1}{x}+ \frac{1}{a- x}\right)$$
$$\int_0^a \frac{dx}{x(a-x)}= \frac{1}{a}\left(\int_0^a \frac{dx}{x}+ \int_0^a\frac{dx}{a-x}\right)$$

$$= \frac{1}{a}\left(ln(x)- ln(a-x)\right)= \frac{1}{a}ln\left(\frac{x}{a-x}\right)$$
and, since that does not exist at x= 0 or x= a, the integral does not exist.