<s>[INST] Calculating the limit $\lim\limits_{a\to0} \int_0^2 {1 \over ax^4+2}\,\mathrm dx$ I have some trouble calculating this integral: $$\lim_{a\to 0} \int_0^2 {1 \over ax^4+2}\,\mathrm dx$$ Got something divided by zero all of the time. 
Thanks in advance for any assistance! 
 [/INST] We can do as hot_queen and Cocopuffs suggest, interchange limit and integration (using suitable justification, such as Lebesgue dominated convergence). This suggests that the result will be $1$.
We can also derive it a bit more elementarily. Note that:
$$\frac{1}{2+ax^4} = \frac12\left(1-\frac{ax^4}{2+ax^4}\right)$$
Hence:
\begin{align}
\int_0^2 \frac{1}{2+ax^4}\,\mathrm dx = \int_0^2 \frac12\,\mathrm dx - \frac12 \int_0^2 \frac{ax^4}{2+ax^4}\,\mathrm dx
\end{align}
and it will suffice to show that:
$$\lim_{a\to0}a \int_0^2 \frac{x^4}{2+ax^4}\,\mathrm dx = 0$$
which, due to the factor $a$, amounts to showing that:
$$\int_0^2\frac{x^4}{2+ax^4}\,\mathrm dx$$
is bounded in a suitable neighbourhood of $a = 0$. So assume that $|a| < 2^{-5}$; then for $0\le x \le 2$, $2+ax^4 \ge \frac32$, so that:
$$0 \le \frac{x^4}{2+ax^4} \le \frac23x^4$$
for $0 \le x \le 2$, which gives bounds on the integral:
$$0 \le \int_0^2 \frac{x^4}{2+ax^4}\,\mathrm dx \le \int_0^2 \frac23x^4\,\mathrm dx = \frac{64}{15}$$
Finally, we can conclude that:
$$0 \le \lim_{a\to0}a\int_0^2 \frac{x^4}{2+ax^4}\,\mathrm dx \le \lim_{a\to0}\frac{64}{15}a = 0$$
and for the original limit, that:
$$\lim_{a\to0}\int_0^2 \frac{1}{2+ax^4}\,\mathrm dx = \lim_{a\to0}\int_0^2 \frac12\,\mathrm dx - \lim_{a\to0}\frac12 \int_0^2 \frac{ax^4}{2+ax^4}\,\mathrm dx = 1 - 0 = 1$$
</s>