Pointwise or Uniform convergence?

Question

Consider the functions $f_n:[-1,1]\to\mathbb{R}$ defined by $$f_n(x):= \frac{x}{\sqrt{x^2 + \tfrac 1n}}$$ and determine whether the convergence is uniform or pointwise.

Answer 1

We have that

If a sequence of functions $(f_n)_{n\geq 1}$ converges pointwise to a function $f$ and uniformly to a function $g$, then $f=g$.

and

If a sequence of continuous functions $(f_n)_{n\geq 1}$ converges uniformly to a function $f$, then $f$ is continuous.

• Prove convergence pointwise to the function $f\colon[-1,1]\mapsto \operatorname{sign}(x)$ as you started: fix $x_0\in[-1,1]$ arbitrarily, and consider the limit of the sequence of real numbers $(f_n(x_0))_{n\geq 1}$. (You may want to consider separately the case $x_0<0$, $x_0=0$ and $x_0>0$.)

• Then, for uniform convergence... see how to apply the two statements above.