$$lim_{n \to \infty} \int_0^1(1 - e^{\frac{-x^2}{n}})x^{-1/2}dx$$
I want to use the Dominated Convergence Theorem to solve this.
Let $f_n = (1 - e^{\frac{-x^2}{n}})x^{-1/2}$.
Step 1: Determining convergence of $f_n$
Fix $x$ to be some constant number.
Now, bringing the limit inside the integral, we have $lim_{n \to \infty}(1 - \frac{1}{e^{\frac{k}{n}}})t$ where $k, t$ are constants.
As $n \to \infty, \ \frac{1}{e^{\frac{k}{n}}} \to 1$
So we get $(1 - 1)t = 0$
So $f_n$ converges to the zero function $f_0$. Note that it doesn't converge at $x = 0$ as we have a division by zero but on $(0, 1]$ it converges to $f_0$ and the measure of $\{0\}$ is $0$ so we have convergence almost everywhere.
Step 2: Determining a dominating function $g$
Now, when $x \in (0, 1)$, for all $n$ we have,
$$e^{\frac{1}{n}} > 1$$
$$\implies e^{\frac{x^2}{n}} \ge 1$$
$$\implies 0 < \frac{1}{e^{\frac{x^2}{n}}} \le 1$$
$$\implies -1 \le - \frac{1}{e^{\frac{x^2}{n}}} < 0$$
$$\implies 0 \le 1 - \frac{1}{e^{\frac{x^2}{n}}} < 1$$
$$\implies \mid 1 - \frac{1}{e^{\frac{x^2}{n}}}\mid < 1$$
$$\implies \mid (1 - \frac{1}{e^{\frac{x^2}{n}}})x^{-1/2}\mid < x^{-1/2}$$
Let $g = x^{-1/2}$
$g$ is integrable as $\int_0^\pi x^{-1/2} = 2\sqrt{x} \mid_0^\pi = 2\sqrt\pi$
So we have a sequence of integrable functions $f_n$ that converges a.e. to the function $f_0$. Let $g = \frac{1}{\sqrt{x}}$. By the dominated convergence theorem as $\mid f_n \mid \le g$ on $(0, 1)$ for all $n$ we have that $$\int_0^1 f_0 = lim_{n \to \infty} \int_0^1 f_n $$
And the integral of the zero function $f_0$ is $0$ so we have that
$$lim_{n \to \infty} \int_0^1(1 - e^{\frac{-x^2}{n}})x^{-1/2}dx = 0$$
Have I made any mistakes here? In particular, I have an issue with the way I brought the limit inside the integral in step 1. I couldn't apply the Monotone Convergence Theorem as we have decreasing sequence, so is there some other way to justify bringing it inside the integral?
