Compute the limit of $\displaystyle\lim_{n \to \infty} \int_0^1 \frac{e^{-nt}}{\sqrt{t}} \, dt$ I am attempting to compute the limit of
$$\displaystyle\lim_{n \to \infty} \int_0^1 \frac{e^{-nt}}{\sqrt{t}} \, dt$$

Proof Attempt
Recall that $\int_a^b f(x) \, dx = \displaystyle\lim_{n \to \infty} \int_a^b f_n(x) \, dx$  provided that  each $f_n$ is continuous and $f_n \to f$ converges uniformly on $[a,b]$ (Thomson & Bruckner, p. 388).
Denote $\langle f_n \rangle$ be a sequence of functions defined by $f_n(x)= \frac{e^{-nt}}{\sqrt{t}}$ where $n \in \mathbb{N}$ and $t \in [0,1]$. Note that each $f_n$ is continuous on $[0,1]$.
Consider, the fact that $f_n \to f$ pointwise, but does not converge to the function $f$ uniformly on $[0,1]$, however. This is because $t=0$ is undefined for the function $f_n \hspace{0.3cm} \forall n \in \mathbb{N}$. But notice that $f_n \to f$ uniformly on $[a,1]$ for $a>0$. So Theorem 9.26 is applicable for $t \in [a,1]$. Thus, consider the following:
$$(A.) \hspace{0.5cm} \displaystyle\lim_{n \to \infty} \int_0^1 f_n(x) = \displaystyle\lim_{n \to \infty} \int_{\frac{1}{n}}^1 f_n(x)$$
since, for $n$ large enough, $\frac{1}{n} < \varepsilon$ for any $\varepsilon > 0$ (Archimedean Property). Since $\displaystyle\lim_{n \to \infty} \langle \frac{1}{n}: n \in \mathbb{N} \rangle = 0$, the above statement $(A.)$ holds.

NOTE: This proof is not complete.
The basic idea of my proof is to observe that $t=0$ breaks the functions $f_n$. So the intuition is to split up $\int f_n \, dt$ into some other integrals that satisfy uniform convergence so that we may use the result from Thomson & Bruckner.
Does this motivation for a proof seem in the right ballpark? Or is there a better means of computing this limit?

You are on the right track! Firstly, I want to note that $$ \lim_{n\to \infty} \int_0^1 \frac{e^{-nt}}{\sqrt{t}}dt$$ does exist. Indeed, on $[0,1]$, $e^{-nt} \geqslant e^{-(n+1)t} $ so $$\int_0^1 \frac{e^{-nt}}{\sqrt{t}}dt $$ is monotone decreasing and bounded from below by $0$.

Now fix $0<\varepsilon<1$ and write $$\int_0^1 \frac{e^{-nt}}{\sqrt{t}}dt = \int_0^\varepsilon \frac{e^{-nt}}{\sqrt{t}}dt +\int_\varepsilon^1 \frac{e^{-nt}}{\sqrt{t}}dt . $$ As you have observed, $f_n $ converges uniformly on $[\varepsilon,1]$. In fact, $f_n \to 0$ uniformly on this interval so $$ \lim_{n \to \infty} \int_\varepsilon^1 \frac{e^{-nt}}{\sqrt{t}}dt = 0 $$ for each $\varepsilon$. Hence, $$\lim_{n \to \infty}\int_0^1 \frac{e^{-nt}}{\sqrt{t}}dt = \lim_{n \to \infty}\int_0^\varepsilon \frac{e^{-nt}}{\sqrt{t}}dt.  $$ Next, $e^{-nt} \leqslant 1$ so $$\int_0^\varepsilon \frac{e^{-nt}}{\sqrt{t}}dt \leqslant \int_0^\varepsilon \frac1{\sqrt{t}}dt =2 \sqrt{\varepsilon} .$$ Thus, $$ 0 \leqslant \lim_{n \to \infty}\int_0^1 \frac{e^{-nt}}{\sqrt{t}}dt \leqslant 2 \sqrt{\varepsilon}. $$ Since this estimate holds for all $\varepsilon>0$ we may take $\varepsilon \to 0^+$ to obtain $$\lim_{n \to \infty}\int_0^1 \frac{e^{-nt}}{\sqrt{t}}dt =0. $$
