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\vspace*{.3in}
\noindent{\bf Beals, page 25, \#2:}
\begin{description}
\item[(a)] Note that for $n,x\geq 0$, we have
$$
e^{nx} = \sum_{j=0}^\infty \frac{(nx)^j}{j!} \leq \frac{n^2x^2}{2}.
$$
Therefore, $n^2 x^2/e^{nx} \leq 2$ for all $n\in\N$ and all $x\in[0,1]$.
In other words, for $x\in[0,1]$ we have $|n^2 x^2/e^{nx}| \leq g(x)$,
where $g(x) := 2\cdot\mathbf{1}_{[0,1]}$.  

Moreover, $\lim_{n\to\infty} n^2 x^2/e^{nx} \to 0$ for every $x\in [0,1]$
(use L'h\^opital's rule, differentiating with respect to $n$, for instance).
So we can apply the dominated convergence theorem to conclude
$$
\lim_{n\to\infty}\int \frac{n^2x^2}{e^{nx}} = 
\int \lim_{n\to\infty}\frac{n^2x^2}{e^{nx}} = \int 0 = 0.
$$
\item[(c)]
Since $\log(1+t) \leq t$ for $t\geq 0$ and $e^s$ is increasing in $s$,
we have
$$
\left(1+\frac{x}{n}\right)^n = e^{n\log\left(1+\frac{x}{n}\right)}
\leq e^{n\cdot(x/n)} = e^x.
$$
Hence for all $n\in\N$,
$$
\left(1+\frac{x}{n}\right)^ne^{-\alpha x} \leq e^{{(1-\alpha)}x},
$$
where the right hand side is integrable on $[0,\infty)$ for $\alpha>1$.
It follows from this and the dominated convergence theorem that
$$
\lim_{n\to\infty}\int_0^\infty \left(1+\frac{x}{n}\right)^ne^{-\alpha x}
= 
\int_0^\infty \lim_{n\to\infty} \left(1+\frac{x}{n}\right)^ne^{-\alpha x}
= 
\int_0^\infty e^{{(1-\alpha)}x}
= 
\frac{1}{\alpha - 1}.
$$
\end{description}
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