So I have to work out $\lim_{n \to \infty} \ \int^2_0 \left(e^x+e^{\frac{x^2}{n}}\right) dx$.
First we will check if the integrand converges normally, we have $x \to x$ as $n \to \infty$ and also $\frac{x^2}{n} \to 0$ as $n \to \infty$, so we then have:
$$e^x + e^\frac{x^2}{n} \to e^x+1 \text{ as } n \to \infty.$$
We can test the convergence of this on our bounds $[0,2]$ by: $$\left|(e^x + e^\frac{x^2}{n}) - (e^x+1)\right| = \left|e^\frac{x^2}{n}-1\right|= \sup_{x \ \epsilon \ [0,2]}\left|e^\frac{x^2}{n}-1\right|≤ \left|e^\frac{4}{n}-1\right| \to 0$$
Which means it does in fact converge uniformly, so for every n the function $e^x+e^\frac{x^2}{n}$ is integrable so we can apply: $$ \int_a^b f = \lim_{n \to \infty} \int_a^b f_n $$
This is where I am stuck, I either don't know where to go from here or I have already made a mistake... Any help would be much appreciated.
Let $g(x) = e^x + e^4$ on $[0,2]$. Then $$f_n(x) = e^x + e^{\frac{x^2}n}\leqslant e^x+e^4=g(x)$$ for all $x\in[0,2]$, and $$\int_0^2 |g(x)|\ \mathsf dx = \int_0^2\left(e^x+e^4\right)\ \mathsf dx = 2e^4 + e^2 - 1<\infty. $$ It follows from the dominated convergence theorem that \begin{align} \lim_{n\to\infty} \int_0^2 f_n(x)\ \mathsf dx &= \int_0^2 \lim_{n\to\infty} f_n(x)\ \mathsf dx\\ &= \int_0^2 (e^x+1)\ \mathsf dx\\ &= e^2 + 1. \end{align}