I would like to show that the following sequence $(f_n)$ of functions converges pointwise to a zero function. We define $f_n(x) = e^{-nx}x^{-1/2}$ for every $x > 0$ and for every natural number $n$.
Here are my attempts.
If $x \geq 1$, then $0 \leq e^{-nx}x^{-1/2} \leq e^{-nx}$ and squeezing both sides with a limit of $n \to \infty$ gives us that $f_n$ converges pointwise a zero function.
The part I struggled with is when $x \in (0,1)$. I am not sure if the following logic works.
I first choose $x \in (0,1)$, then $x$ is no longer a variable but a fixed value, therefore we can safely say that $$\lim_{n\to\infty}e^{-nx}x^{-1/2} = 0$$
Therefore $(f_n)$ converges pointwise to a zero function.
Is my reasoning logical? If not, then could someone show me how to prove this? Any help will be greatly appreciated.
Best Answer
You can always apply the idea that, for each single $x$, $x$ is a constant. So, since $\lim_{n\to\infty}e^{-nx}=0$, you also have $\lim_{n\to\infty}e^{-nx}x^{-1/2}=0$. Note that you do not need the squeeze theorem at all.