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! 
 A: 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:
$$\lim_{a\to0}\int_0^2 \frac{1}{2+ax^4}\,\mathrm dx = \lim_{a\to0}\int_0^2 \frac12\,\mathrm dx - \lim_{a\to0}\frac12 \int_0^2 \frac{ax^4}{2+ax^4}\,\mathrm dx = 1 - 0 = 1$$
A: You shouldn't be calculating any antiderivative of $(ax^4 + 2)^{-1}$. Instead, use the dominated convergence theorem and calculate $$\int_0^2 \lim_{a \rightarrow 0} \frac{1}{ax^4 + 2} \, \mathrm{d}x.$$
A: It is obvious that
$$
I(a)=\int_0^2\frac{1}{ax^4+2}\,dx
$$
converges for $a > -\frac18$ and diverges for $a \le -\frac18$.
Now, for every $a >-\frac18$ we have
\begin{eqnarray}
|I(a)-1|&=&|I(a)-I(0)|=\left|\int_0^2\left(\frac{1}{ax^4+2}-\frac12\right)\,dx\right|=\frac12\left|\int_0^2\frac{ax^4}{ax^4+2}\,dx\right|\\
&\le&\frac12\int_0^2\frac{|a|x^4}{|ax^4+2|}\,dx \le \frac{|a|}{4\min\{1,1+8a\}}\int_0^2x^4\,dx\\
&=&\frac{8|a|}{5\min\{1,1+8a\}}.
\end{eqnarray}
Since
$$
\lim_{a \to 0}\frac{8|a|}{5\min\{1,1+8a\}}=0,
$$ 
it follows that
$$
\lim_{a\to 0} I(a)=I(0)=1.
$$
