Limit of integral, show limit exists and compute it Suppose $f : [0,\infty) \rightarrow \mathbb{R}$ is continuous and such that $|f(x)| \leq 1 + x^{2}, \forall x \geq 0$. Then how do I show that: $L = \lim_{n \rightarrow \infty} \int_{0}^{n} \frac{f(x/n)}{(1+x)^{4}} dx$ exists and how do I compute it? If I give a bound for it, I obtain that $|L| \leq \lim \int_{0}^{n} \frac{1+(x/n)^{2}}{(1+x)^{4}} dx = \lim_{n \rightarrow \infty} \frac{1}{3} - \frac{1}{3(n+1)^{3}} + \frac{-3n^{2} - 3n - 1 + (n+1)^{3} }{3n^{2}(n+1)^{3}}$. Taking the limit as $n$ goes to infinity gives that $|L| \leq \frac{1}{3}$. However, this bound does not hold for all $f(x)$, as $f(x) = 0$ gives $L = 0$. I am assuming that the limit depends on $f$, but I don't know how to proceed. 
 A: Take $g_n(x)=\frac{f(\frac{x}{n})}{(x+1)^4}1_{[0,n]}(x)$
Note that $$|g_n(x)| \leq \frac{1+x^2}{(x+1)^4}:=h(x),\forall n \in \Bbb{N}$$ and  $h(x) \in L^1[0,+\infty)$
Also $g_n(x) \to \frac{f(0)}{(x+1)^4}$ pointwise on $[0,+\infty)$
So by Dominated Convergence theorem, the limit is $$L=f(0)\int_0^{\infty}\frac{1}{(x+1)^4}dx=\frac{f(0)}{3}$$
A: By substitution $x=nt$ the integral in question reduces to $$\int_{0}^{1}\frac{nf(t)}{(1+nt)^4}\,dt$$ and we have $$\int_{0}^{1}\frac{n}{(1+nt)^4}\,dt=\frac{1}{3}-\frac{1}{3(1+n)^3}\to \frac{1}{3}$$ By continuity of $f$ we have for every $\epsilon >0$ a $0<\delta<1$ such that $$|f(t) - f(0)|<\epsilon$$ whenever $0\leq t\leq \delta$. Further $|f|$ is bounded on $[0,1]$ by a positive constant $M$. We have $$A_n=\left|\int_{0}^{1}\frac{nf(t)}{(1+nt)^4}\,dt-\frac{f(0)}{3}\right|=\left|\int_{0}^{1}\frac{n(f(t)-f(0))}{(1+nt)^{4}}\,dt-\frac{f(0)}{3(1+n)^3}\right|$$ and by triangle inequality the above does not exceed $$\int_{0}^{\delta}\frac{n|f(t)-f(0)|}{(1+nt)^4}\,dt+\int_{\delta}^{1}\frac{n|f(t)-f(0)|}{(1+nt)^{4}}\,dt+\frac{M} {n^3}$$ and this is clearly less than $$\frac{\epsilon} {3}+\frac{2M}{3(1+n\delta)^3}+\frac {M} {n^3}$$ Since the above tends to $\epsilon/3$ as $n\to\infty $ it follows that $$0\leq \liminf_{n\to\infty} A_n\leq\limsup_{n\to \infty} A_n\leq\frac{\epsilon} {3}$$  The above inequality holds for every $\epsilon>0$ and therefore $A_n\to 0$. It follows that the desired limit in question is $f(0)/3$.

The above argument requires that $f$ be Riemann integrable on $[0,1]$ and that $f(x) \to L$ when $x\to 0^{+}$. The desired limit is then $L/3$. Other hypotheses of the question are not really necessary. 
