Fix some $\delta\in \mathbb R$ and let $f:[0,\infty)\rightarrow \mathbb R$ be given by the equation $$f(x)=\frac{\sin(x^2)}{x}+\frac{\delta x}{1+x}$$
Show that,   
$\lim_{n\rightarrow\infty}\int_{0}^{a}f(nx)\ dx=a\delta$
for each $\ a>0$.
My attempt:
$\lim_{n\rightarrow\infty}\ f(nx)=\delta$ and $|f(nx)|\le (\frac{1}{nx}+\delta)$ 
then the right-hand-side of the inequality is integrable
but what if $a=\infty$ ?
 A: Fortunately, $\infty$ is not a number. So saying that $a > 0$ implies that $a \in \mathbb{R}$, and is therefore finite.
However, if the upper bound of the integral were $\infty$, it is easy to show that $\displaystyle \lim \int_0^\infty f(nx) \mathrm{d}x \to \infty$, so that in the looser sense it is true that $\displaystyle \lim_{n\rightarrow\infty}\int_{0}^{a}f(nx)\ dx=a\delta$, where both sides are 'infinite' if $a$ is 'infinite.'
A: You have $\int_0^a f(nx) dx = \frac{1}{n} \int_0^{na} f(t) dt$.
It is straightforward to find that $ \int_0^{na} \frac{x}{1+x} dx = n a -\log(n a+1)$, and so $\lim_{n \to \infty} \frac{1}{n} \int_0^{na} \frac{x}{1+x} dx  = a$.
Note that $\frac{\sin x^2}{x} \le \begin{cases} 1, & x \in (0,1] \\
\frac{1}{x}, & 1 <x \end{cases}$. Hence for $n$ sufficiently large, we have
$\int_0^{n a} \frac{\sin x^2}{x} dx \le \int_0^1 dx + \int_1^{n a} \frac{1}{x}dx = 1+\log(n \alpha)$, hence $\lim_{n \to \infty} \frac{1}{n} \int_0^{na} \frac{\sin x^2}{x} dx = 0$
Combining gives  the desired result.
