Does $f_n(x) = \frac{x}{1 + nx^2}$ converge uniformly for $x \in \mathbb{R}$?

Question

Does $f_n(x) = \frac{x}{1 + nx^2}$ converge uniformly to $f(x)$ for $x \in \mathbb{R}$?

$\lim_{n \rightarrow \infty} f_n = f(x) = 0$

For $f_n(x) \rightarrow f(x)$ uniformly, I have to show that $sup_{x \in \mathbb{R}} |f_n(x) - f(x)| \rightarrow 0$

$sup_{x \in \mathbb{R}} |\frac{x}{1 + nx^2}| = \frac{1}{1 + n}$???

So I was thinking, $x = 1$ will give me the largest value, so $lim_{n \rightarrow \infty} \frac{1}{1 + n} = 0$, so that would mean $f_n(x) \rightarrow f(x)$ uniformly, right? However, the answer says that $f_n(x)$ does not converge uniformly to $f(x)$. Why is this?

Answer 1

We have $$\mathop {\sup }\limits_{x \in \mathbb{R}} \left| {{f_n}\left( x \right) - 0} \right| = \mathop {\sup }\limits_{x \in \mathbb{R}} \left| {\frac{x}{{1 + n{x^2}}}} \right| = {\left( {\left| {\frac{x}{{1 + n{x^2}}}} \right|} \right)_{x = 1/\sqrt n }} = \frac{1} {{2\sqrt n }} \to 0.$$ So, $f_n(x)$ converge uniformly to $f(x)=0$.

Answer 2

$\underset{n\rightarrow\infty}{\lim} \sup_{x\in \mathbb{R}} |f_n(x) - f(x)| = \underset{n\rightarrow\infty}{\lim} \sup_{x\in \mathbb{R}} |\frac{x}{1+nx^2} - 0| = \underset{n\rightarrow\infty}{\lim} |\frac{1}{2\sqrt n}| = 0$, therefore $f_n$ does converge uniformly.