Interchanging limit and integral in $\lim\limits_{n\to\infty} \int_{0}^{a}\frac{e^x}{1+x^n}\mathrm dx$ I am attempting the following question. As per the solution provided, the limit and the integral have been interchanged, but I do not see how that would go about. I have not yet studied DCT or MCT. So is it possible to intuitively study this interchange using properties of the definite integral.
$$\lim_{n\to \infty}\int_{0}^{a}\frac{e^x}{1+x^n}\,\mathrm dx$$
Any hints are appreciated. Thanks.
 A: $\displaystyle \left| \frac {e^{x}} {1+x^{n}}-e^{x}\right|=e^{x} \frac {x^{n}} {1+x^{n}} \leq e x^{n}$. So $\displaystyle \left|\int_0^{1} \left[\frac {e^{x}} {1+x^{n}}-e^{x}\right] \mathrm{d}x\right|\leq e\int_0^{1}x^{n}\mathrm{d}x=\frac e {n+1} \to 0$. This implies that $\displaystyle \lim \int_0^{1} \frac {e^{x}}{1+x^{n}}=e-1$.  If $a\leq 1$ then you can easily modify this argument to see that the given limit is $e^{a}-1$.
Now let $a>1$.
We shall show that $$\lim \int_1^{a} \frac {e^{x}} {1+x^{n}} \mathrm{d}x =0.$$
Let $\epsilon >0$ and choose $\delta >0$ such that $\displaystyle \int_1^{1+\delta} \frac {e^{x}} {1+x^{n}} \mathrm{d}x \leq e^{a} \int_1^{1+\delta}  \mathrm{d}x=e^{a}\delta <\epsilon$. Next note that $$\int_{1+\delta} ^{a} \frac {e^{x}} {1+x^{n}} \mathrm{d}x \leq e^{a} \int_{1+\delta} ^{a}\frac  1 {x^{n}} \mathrm{d}x=\frac  1 {1-n} e^{a}(a^{1-n}-(1+\delta)^{n-1}) \to 0.$$
It should now be clear that limit of the integral is equal to the integral of the limit.
A: Here is an approach using dominated convergence theorem. I'm not sure a more elementary solution exists (although you ask for this), but at least this gives an idea what the answer should be.
For $x \in [0,a]$ and $n \geq 1$, we have
$$\left|\frac{e^x}{1+x^n}\right| \le e^x$$
and $$(x \mapsto e^x) \in \mathcal{L}^1([0,a])$$
Moreover, for all $x \in [0,1[$ we have
$$\lim_n \frac{e^x}{1+x^n}= e^x$$
and for $x \in ]1,a]$ we have
$$\lim_n \frac{e^x}{1+x^n}= 0$$
Hence, dominated convergence applies but there are multiple cases to consider.
Suppose first $a > 1$. Then
$$\lim_n \int_0^a \frac{e^x}{1+x^n} dx = \int_0^1 e^xdx + \int_1^a 0 dx = e-1$$
If $a \leq 1$, then
$$\lim_n \int_0^a \frac{e^x}{1+x^n} dx = \int_0^a e^xdx = e^a -1$$
