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?

When $\epsilon$ is very small you get something which looks like 
$$\int_0^1 1~ dx=1$$
so just try to see if the difference between your integral and $1$ tends to $0$ as $\epsilon\to 0$. You have 
\begin{align} 
\left|\int_0^1 \frac{1}{1+\epsilon x^3} dx -1\right| &= \left| \int_0^1 \frac{1}{1+\epsilon x^3} dx -\int_0^1 1~dx \right| \\
&= \left|\int_0^1 \frac{-\epsilon x^3}{1+\epsilon x^3} dx \right| \\ 
&\leq \int_0^1 \frac{\vert\epsilon x^3\vert}{\vert1+\epsilon x^3\vert} dx \\
&\leq \epsilon.
\end{align}
So $$\int_0^1 \frac{1}{1+\epsilon x^3} dx\longrightarrow1\quad\text{as}\quad \epsilon\to 0^+ . $$
