Pointwise convergence of the sequence of functions

Question

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.

Answer 1

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.

Answer 2

Yes, your reasoning is logical. But still overly complicated, since the argument that $x$ is viewed as fixed can be used just the same for $x\ge1$, and hence it is not necessary to distinguish two cases.

Answer 3

It's been a long time since the last chance I had to study functional sequences, so correct me if I wrong in the following:

Let $f$ be a function defined in a domain $D$. We say that $(f_n)_{n \in \mathbb{N}}$ converges (point-wise) to $0$ if, and only if:

$$\forall x \in D, \forall \epsilon > 0, \exists \eta \in \mathbb{N}: k>\eta \implies |f_k(x)| <\epsilon$$

With this sentence being equivalent to say:

$$\forall x \in D, \lim_{n \to \infty} f(x)=0$$

So there you have it! Exactly as you've done it for the $(0,1)$ interval. And it would have worked in all the domain! In short, pointwise-convergence is just about taking a bunch of limits $n \to \infty$ for the different values of $x$. Uniform convergence is a more complicated story, though