Is the convergence pointwise or uniform?

Question

For the sequence of functions, $\{f_n\}$ $n>-1$

Let $f_n:[0,\infty)\to \mathbb R$, and let it be defined by $f_n(x)=\frac{x}{1+n+x}$.

(1) Does $f=\lim_{n\to \infty} f_n$ exist?

(2) If this limit does indeed exist, does $f_n(x)$ converge point wise or uniformly on $[0,\infty)$ ?

I know how to do this for $\sqrt(x)$ , but this one seems more complicated to me. any help is greatly appreciated.

Answer 1

The first one is pretty simple, you just have to calculate $\lim_{n \to \infty} \frac{x}{1+n+x}$. Keeping in mind that $x$ is fixed as you take the limit, this shouldn't be hard to do. You should find that the limit is the zero function.

The idea in the second part is to actually quantify the convergence rate that we just described. We've already identified that the limit is $0$, so $|f_n(x)-f(x)|=\frac{x}{1+n+x}$. So the relevant convergence inequality is $\frac{x}{1+n+x} < \varepsilon$. Now solve this inequality for $n$. You should get something like $n>g(x,\varepsilon)$ for some $g$. Try to see whether you can write $g(x,\varepsilon) \leq h(\varepsilon)$ for some $h$, or not. If you can, then you've just proven uniform convergence; if you can't, then because all of your steps were reversible, you've just proven that you don't have uniform convergence.

Answer 2

Hints:

1. For fixed $x \in [0, \infty)$, what is $\lim_{n\to \infty} (1+n+x)$? What does that mean for $\lim_{n\to \infty} \frac x {1+n+x}$?

2. For fixed $n \in \mathbb N$, what is $\lim_{x\to \infty} \frac x {1+n+x}$? Then recall the definition of uniform convergence.