What is wrong with my application of Lebesgue Dominated Convergence Theorem in these two examples? Background
I seem to be having issues recognizing valid bounding functions when applying the Lebesgue Dominated convergence theorem. Here are two examples I did that I do not think are justified.
Example 1: Show that $$\lim_{n\to\infty} \int_0^\infty ne^{-nx}\sin(1/x) \ dx$$ exists and determine its value.
Solution: Let $u=nx$. Then we may rewrite the integral as $$\int_0^\infty e^{-u}\sin\left(\frac{n}{u}\right) \ du.$$
Note that $$\bigg\lvert e^{-u}\sin\left(\frac{n}{u}\right)\bigg\rvert\leq e^{-u},$$ and since $e^{-u}$ is integrable, the LDCT applies. The limit is then $0$.
Example 2: Let $g:\mathbb{R} \rightarrow \mathbb{R}$ be integrable and let $f:\mathbb{R} \rightarrow \mathbb{R}$ be bounded, measurable, and continuous at $1$. prove that $$\lim_{n\to\infty}\int_{-n}^nf\left(1+\frac{x}{n^2}\right)g(x) \ dx$$ exists and determine its value. 
Solution: $f$ is bounded so for all $x$,$|f|\leq K$ for some $k \in \mathbb{R}$. Thus $$\bigg\lvert f\left(1+\frac{x}{n^2}\right)g(x) \chi_{[-n,n]}\bigg\rvert\leq k|g(x)|.$$ Since $k|g(x)|$ is integrable, we again have by the LDCT that the integral converges. The limiting value is $\int_\mathbb{R}f(1)g(x) \ dx.$
My Question
I am not really looking for the solution to these problems. Rather I wish to know how I am applying the LDCT wrong in both of these (I am fairly sure that I am applying it wrong).
 A: In Example 1, after your substitution, the integral should be
$$
\lim_{n\to\infty}\int_0^\infty e^{-u}\sin\left(\frac nu\right)\,\mathrm{d}u
$$
Indeed, the integrand is dominated by $e^{-u}$, but it doesn't converge pointwise to $0$.
Let's try integrating by parts, then substitute $u=nx$
$$
\begin{align}
\int_0^\infty ne^{-nx}\sin\left(\frac1x\right)\,\mathrm{d}x
&=\int_0^\infty nx^2e^{-nx}\,\mathrm{d}\cos\left(\frac1x\right)\\
&=\int_0^\infty\cos\left(\frac1x\right)(n^2x^2-2nx)e^{-nx}\,\mathrm{d}x\\
&=\int_0^\infty\frac1n\cos\left(\frac nu\right)(u^2-2u)e^{-u}\,\mathrm{d}u
\end{align}
$$
Now, the integrand is bounded by $|u^2-2u|\,e^{-u}$, which is integrable, and the integrand converges pointwise to $0$.
In fact, since $\int_0^\infty|u^2-2u|\,e^{-u}\,\mathrm{d}u=\frac8{e^2}$, we have
$$
\left|\,\int_0^\infty ne^{-nx}\sin\left(\frac1x\right)\,\mathrm{d}x\,\right|\le\frac8{ne^2}
$$
A: The second one seems fine to me.
For the first, you simply hid the dependency on $n$.
That is, it's fine to set $u = nx$ inside the limit; for you get that for every fixed $n$, your integrand is bounded by $e^{-u}$. But anyway your $u$ depends on $n$; so what you've shown is that the $n$-th integrand can be bound by a function that depends on $n$, and this is not what the LDCT requires.
A: The mistake in the first problem is very subtle: you first note that $f_n \to 0$ pointwisely; after the change of variables you get some new functions such that $|g_n| \le \Bbb e ^{-u}$, but how do you connect the $f$s and the $g$s? Plus, $g_n \not\to 0$.
The second one is fine.
