Cfrgtkky

Let $f_n(x)=begin{cases}frac{e^{x^2}}{x^2} &xinleft[1/n,1right]\ 0 &xin(-infty,1/n)cup(1,+infty)end{cases}.$

This converges to $f(x)=begin{cases}frac{e^{x^2}}{x^2} &xinleft(0,1right]\ 0 &xin(-infty,0]cup(1,+infty)end{cases}$ pointwise on $Bbb R$ and uniformly on $(-infty,-varepsilon]cup[varepsilon,+infty)$ for every $varepsilon>0$.

I'm then asked to find $lim_{ntoinfty}int_0^1f_n(x)dx$, and I think the following holds: $$lim_{ntoinfty}int_0^1f_n(x)dx=lim_{ntoinfty}int_0^{1/n}0+lim_{ntoinfty}int_{1/n}^1frac{e^{x^2}}{x^2}dx=lim_{ntoinfty}int_{1/n}^1frac{e^{x^2}}{x^2}=int_{0}^1frac{e^{x^2}}{x^2}=+infty.$$Am I correct?

You are correct. Also, you can use Dominated convergence Theorem with the Dominant function $f$ observing that $|f_n(x)|leq |f(x)|$ $forall n$ $forall x$ since $$frac{e^{x^2}}{x^2}geq 0 hspace{2mm}forall xin (0,1].$$ Thus, by the Dominated convergence theorem,

$$lim_{nrightarrow infty}int f_n=int lim_{nrightarrow infty}f_n =int f =int_{(0,1]}frac{e^{x^2}}{x^2}.$$

Since $$ +infty=int_{(0,1]}frac{1}{x^2}leq int_{(0,1]}frac{e^{x^2}}{x^2}, $$

we have $$lim_{nrightarrow infty}int f_n=int_{(0,1]}frac{e^{x^2}}{x^2}=+infty.$$

Yes it is true. This follows from Beppo Levi's monotone convergence theorem for Lebesgue integral here: https://en.wikipedia.org/wiki/Monotone_convergence_theorem

$f_n(x) ge frac{1}{x^2}$ on $[1/n,1]$ and by positivity of Riemann integral
$$ int_0^1 f_n(x) dx = int_{1/n}^1 frac{e^{x^2}}{x^2} dx ge int_{1/n}^1 frac{1}{x^2} dx = n-1.$$

The result follows.