Find $\lim_{n \to \infty} n^{\alpha} \int_{n}^{\infty} \frac{f(x/n^2)}{x^{\alpha + 1}}(x-n)dx$ I am looking at an old exam in my measure theory and integration class.
I am trying to solve a problem and am wondering if I am doing it right. 
Problem
Let $f$ be a bounded measurable function on $\mathbb R_+$ with $f(0)=0$ and $f$ differentiable from the right in $0$ with $f'(0)=1$. Find where $\alpha > 2$
$$ \lim_{n \to \infty}  n^{\alpha} \int_{n}^{\infty} \frac{f(x/n^2)}{x^{\alpha + 1}}(x-n)dx$$
Attempt at solution
I tried using the substitution $u=x/n^2$ and got the integral 
$$\lim_{n \to \infty} \int_{1/n}^{\infty} \frac{n^{\alpha}f(u)n^2(un^2-n)du}{u^{\alpha+1}n^{2\alpha+2}}=\lim_{n \to \infty} \int_{1/n}^{\infty} \frac{f(u)(un^2-n)du}{u^{\alpha+1}n^{\alpha}}$$ $$=
\lim_{n \to \infty} \int_{1/n}^{\infty}( \frac{f(u)}{u^{\alpha}n^{\alpha-2}}-\frac{f(u)}{u^{\alpha+1}n^{\alpha-1}})du=\lim_{n \to \infty} \int_{0}^{\infty}( \frac{f(u)}{u^{\alpha}n^{\alpha-2}}-\frac{f(u)}{u^{\alpha+1}n^{\alpha-1}})\mathbb{1}_{[1/n,\infty)}du$$ 
Now I split the integral and note that $$\frac{f(u)}{u^{\alpha}n^{\alpha-2}}\mathbb{1}_{[1/n,\infty)} \leq \frac{M}{u^\alpha}[1/n,\infty)$$ since $f$ is bounded. And the same for the other part. Then I would use the dominated convergence theorem to move the limit under the integral sign. Am I understanding this correctly? Do I not have to do something with the fact that $f'(0)=1$?
Thank you!
 A: I think a different substitution's probably the way forward: try instead $x=n(1+u)$, so $dx = n \, du$. Then if the contents of the limit is $I_n$,
$$ I_n = \int_0^{\infty} \frac{n^{\alpha}}{(n(1+u))^{\alpha+1}} f\left( \frac{1+u}{n} \right) (nu) n \, du, $$
which cancels down to
$$ I_n = \int_0^{\infty} \frac{u}{(1+u)^{\alpha}} \left[ f\left( \frac{1+u}{n} \right) \frac{n}{1+u} \right] \, du, $$
and we have to do something with the square bracket. We guess we want it to look like a derivative quotient. We have, using the properties in the question,
$$ f\left( \frac{1+u}{n} \right) \frac{n}{1+u} = \frac{f((1+u)/n)-f(0)}{(1+u)/n}, $$
which has limit $f'(0)=1$ as $n \to \infty$.
$f$ is bounded by $M$, say, so the integrand is bounded by $Mu(1+u)^{-\alpha}$, which is integrable since $\alpha>2$. Applying the Dominated Convergence Theorem,
$$ \lim_{n \to \infty} I_n = \int_0^{\infty} \frac{u}{(1+u)^{\alpha}} \left[ \lim_{n \to \infty} \frac{f((1+u)/n)-f(0)}{(1+u)/n}\right] \, du = \int_0^{\infty} \frac{u}{(1+u)^{\alpha}} \, du,  $$
which is easy enough to evaluate using integration by parts; the final answer is
$$ \frac{1}{(\alpha-1)(\alpha-2)}. $$
