It's an exercise I took from an exam of Measure Theory.
Using all the theorems it needs, evaluate: $$\lim_{n \rightarrow \infty} \int_0^{+\infty} \frac{e^{-n^2x}}{\sqrt{|x-n^2|}} dx $$
It has an additional hint: from $n>2$ look at the integral like $$ \int_0^{+\infty} = \int_0^1 + \int_1^{n^2-n} + \int_{n^2-n}^{n^2+n}+\int_{n^2+n}^{+\infty}$$
Moreover the function has no elementary primitive, but it is the product of two functions that do have elementary primitive. This can be helpful after having done the right considerations.
Following the given hint we have that for $n\geq 2$,
i) if $x\in [0,1]$ then $$\frac{e^{-n^2x}}{\sqrt{|x-n^2|}}\leq \frac{1}{\sqrt{n^2-1}},$$ ii) if $x\in [1,n^2-n]$ then $$\frac{e^{-n^2x}}{\sqrt{|x-n^2|}}\leq \frac{e^{-n^2}}{\sqrt{n}},$$ if $x\in [n^2+n,+\infty)$ then $$\frac{e^{-n^2x}}{\sqrt{|x-n^2|}}\leq \frac{e^{-n^2x}}{\sqrt{n}}.$$ By using these estimates, you may show that the three related integrals go to zero. It remains to consider the integral: $$\int_{n^2-n}^{n^2+n}\frac{e^{-n^2x}}{\sqrt{|x-n^2|}} \,dx\leq e^{-n^2(n^2-n)}\int_{n^2-n}^{n^2+n}\frac{dx}{\sqrt{|x-n^2|}}=e^{-n^2(n^2-n)}2\int_{0}^{n}\frac{dt}{\sqrt{t}}.$$ Can you take it from here?