Showing that $\lim_{k\rightarrow 0}\int_0^1\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}} = \int_0^1\frac{dx}{\sqrt{(1-x^2)}}$ So I'm trying to show that:
$$\lim_{k\rightarrow 0}\int_0^1\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}} = \int_0^1\frac{dx}{\sqrt{(1-x^2)}}$$
I guess this boils down to a solid understanding of uniform convergence.  
There's also the second issue that the theorem in Rudin's PMA which discusses the exchange of a definite integral and a limit says that the the sequence of integrands must converge to the limit integrand uniformly on a closed interval, in this case that would be $[0,1]$.  But of course our function is only defined on $[0,1)$, thus should I be considering the interval $[0,1-\epsilon]$?
As far as proving uniform convergence, I was looking at the the sequence:
$$M_{a_n} = \sup_{x\in [0,1-\epsilon]}|\frac{1}{\sqrt{(1-x^2)(1-(a_n)^2x^2)}} - \frac{1}{\sqrt{(1-x^2)}}|$$
for any sequence $a_n\rightarrow 0$.
And trying to prove that the sequence $M_{a_n}$ goes to zero, but it doesn't seem to upon pulling out the factor $\frac{1}{\sqrt{1-x^2}}$.  
To sum up, this problem with its complexities is just a bit beyond my level of understanding and comfort with analysis.  Could someone help me sort through it?  Thanks.

What you want to show is that 
$$\frac{1}{\sqrt{(1-x^2)(1-k^2x^2)}}\to \frac{1}{\sqrt{1-x^2}}$$
on uniformly $[0,1-\epsilon]$ for all $\epsilon>0$ and that 
$$\lim\limits_{\epsilon\to 0}\int_0^{1-\epsilon}\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}\to \int_0^{1}\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}$$ uniformly in $k$. The first fact gives us
$$\lim\limits_{k\to 0}\int_0^{1-\epsilon}\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}=\int_0^{1-\epsilon}\frac{dx}{\sqrt{1-x^2}}$$
for all $\epsilon>0$, and since integration is continuous
$$\lim_{\epsilon\to 0}\int_0^{1-\epsilon}\frac{dx}{\sqrt{1-x^2}}=\int_0^{1}\frac{dx}{\sqrt{1-x^2}}$$
while adding the second fact gives 
$$\lim\limits_{\epsilon\to 0}\lim\limits_{k\to 0}\int_0^{1-\epsilon}\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}=\lim\limits_{k\to 0}\int_0^{1}\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}$$
and from the these three equalities we can conclude the desired result.
To prove the first fact, note that
$$\begin{align}
\sup_{x\in [0,1-\epsilon]}\left|\frac{1}{\sqrt{(1-x^2)(1-k^2x^2)}}-\frac{1}{\sqrt{1-x^2}}\right| &\le\left(\sup_{x\in [0,1-\epsilon]}\frac{1}{\sqrt{1-x^2}}\right)\sup_{x\in [0,1-\epsilon]}\left|\frac{1}{\sqrt{1-k^2x^2}}-1\right|\\
&\le\left(\sup_{x\in [0,1-\epsilon]}\frac{1}{\sqrt{1-x^2}}\right)\left|\frac{1}{\sqrt{1-k^2(1-\epsilon)^2}}-1\right|\to 0\\
\end{align}$$
giving uniform convergence. For the second fact, note that
$$\int_{1-\epsilon}^1\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}\le \frac{1}{\sqrt{1-k^2}}\int_{1-\epsilon}^1\frac{dx}{\sqrt{1-x^2}}=\frac{1}{\sqrt{1-k^2}}\int_{0}^1\chi_{[1-\epsilon,1]}(x)\frac{dx}{\sqrt{1-x^2}}$$
where $\chi_{[1-\epsilon,1]}$ is the indicator function for the interval $[1-\epsilon,1]$. The function $\chi_{[1-\epsilon,1]}(x)\frac{1}{\sqrt{1-x^2}}$ converges pointwise to $0$ except at $x=1$ and is dominated by $\frac{1}{\sqrt{1-x^2}}$ which has finite integral, thus by the Dominated Convergence Theorem we see that $\int_{1-\epsilon}^1\frac{dx}{\sqrt{1-x^2}}\to 0$. If we bound $k$ below $1$ (say $k\le 1/2$) we see that
$$\frac{1}{\sqrt{1-k^2}}\int_{1-\epsilon}^1\frac{dx}{\sqrt{1-x^2}}\leq \frac{4}{3}\int_{1-\epsilon}^1\frac{dx}{\sqrt{1-x^2}}\to 0$$
which is clearly uniform in $k$.
