Calculate $\lim_{\epsilon\rightarrow 0^+} \int_0^1 \frac{1}{\epsilon x^3+1} dx$ 
Calculate $\lim_{\epsilon\rightarrow 0^+} \int_0^1 \frac{1}{\epsilon x^3+1} dx$

I tried to use:
$$\int_0^1 f(x) \le \int_0^1 \frac{1}{1+\epsilon x^3} dx \le \int_0^1 \frac{1}{1+0} dx=1$$However I have a problem to find $f(x)$ such that $\int_0^1 f(x) \rightarrow 1$ because when I will take $f(x)=\frac{1}{1+a}, a > 0$ then I have $\int_0^1 f(x)=\frac{1}{1+a}$. Have you got any ideas?

Since $\epsilon>0,$ you have $f_\epsilon (x)\leq f(x),$ where $f_\epsilon(x)=\frac{1}{1+\epsilon x^3}$ and $f(x)=1.$ By dominated convergence theorem ($f$ is integrable on $[0,1]$), $f_\epsilon(x)\to f(x)$ as $\epsilon \to 0^+$ so $\int_0^1 f_\epsilon(x)\,\mathrm{d}x\to \int_0^1 f(x)\,\mathrm{d}x=1.$
