How to prove the non uniform convergence of sequence of functions.

Question

The sequence of functions $\{f_n\}$, defined by $$f_n(x) = n\log \left(1+\frac{x^2}{n}\right)$$ is not uniformly convergent on $\mathbb{R}$.

We have $\lim_\limits{n\to \infty}f_n(0)= 0$ and $$ \lim_{n\to \infty}f_n(x)= \lim_{n\to \infty} \frac{\left(\frac{1}{1+\frac{x^2}{n}}\right)}{\left(-\frac{1}{n^2}\right)} \times \frac{-x^2}{n^2} = x^2. $$ How to proceed now?

Answer 1

Uniform convergence would mean $$\sup\limits_{x \in \mathbb{R}} \left|n\log \big(1+\frac{x^2}{n}\big) - x^2\right| \to 0$$ as $n \to \infty$. To show that this is not the case, choose some sequence $(x_n)_{n \in \mathbb{N}}$ such that $$\left|n\log \big(1+\frac{x_n^2}{n}\big) - x_n^2\right| \not\to 0,$$ e.g. $x_n = \sqrt{n}$ should work.

Answer 2

Just recall the definition.

The sequence $(f_n)_{n\in\mathbb{N}}$ is uniformly convergent to $f$ on a closed set $A$ if $\sup_A |f_n(x)-f(x)|\rightarrow 0$ as $n\rightarrow +\infty$.

Thus you have to show that $\sup_{\mathbb{R}} \left|n\log\left(1+\frac{x^2}n\right)-x^2\right|$ does not tend to $0$.