Evaluate the limits $$\lim_{p\rightarrow\infty}\int_0^1e^{-px}(\cos x)^2\text{d}x$$
I tried to prove the integrand is uniformly convergent so that the limit and integral can be exchanged. But I failed. Any ideas?
 A: There is no need for uniform convergence; using dominated convergence will do the job just fine. In fact, this is not uniform convergence at all (remember that the uniform convergence of a continuous function is continuous, but this is not the case, as the function it converges to has a discontinuity at $x=0$).
Also, if you cannot use dominated convergence, even the squeeze theorem will do. We have
$$\begin{align}
0 \leq \int\limits_{0}^{1} e^{-px}(\cos(x))^{2}\;\mathrm{d}x &\leq \int\limits_{0}^{1} e^{-px}\;\mathrm{d}x \\
&=\left. \frac{-e^{-px}}{p} \right|_{x=0}^{1} \\[5pt]
&=\frac{1}{p} - \frac{e^{-p}}{p}
\end{align}$$
Now $\lim\limits_{p\rightarrow\infty}\frac{e^{-p}}{p}=0$ using L'Hopital, so this does indeed squeeze in.
A: This is actually a simple integration by parts question and you do not necessarily need to invoke any fancy theorems. The integral can be evaluated by elementary means. Use $\cos^2 x = [1 + \cos(2x)]/2$. Then integrate $e^{-px} \cos(2x)$ by parts (twice). I do not think you should have any trouble integrating $e^{-px}$. Lastly, take the limit as $p \to \infty$.
A: Let $\epsilon\gt 0$. Then there is a $P$ such that if $p\gt P$, and $x\gt \epsilon/2$, then $e^{-px}\lt \epsilon/2$.  So for $p\gt P$, our integral is $\lt (1)(\epsilon/2)+(\epsilon/2)(1-\epsilon/2)$, which is $\lt \epsilon$.
