I am given that:
For $n \in \mathbb{N}$, define $f_n: \mathbb{R} \to \mathbb{R}$ by
$$f_n(x)=\frac{x^{4n}}{4+x^{4n}}.$$
I need to determine whether the sequence $(f_n)$ converges uniformly on $\mathbb{R}$.
This is what I have done:
\begin{align} \lim_{n \to \infty}f_n(x)&=\lim_{n \to \infty}\frac{x^{4n}}{4+x^{4n}}= \begin{cases} 0, & \text{if}\ |x|<1 \\ 1, & \text{if}\ |x|\geq1 \end{cases}\\ \\ &\implies f(x)=\begin{cases} 0, & \text{if}\ |x|<1 \\ 1, & \text{if}\ |x|\geq1 \end{cases} \end{align}
So the first case would be when $|x|<1$:
\begin{align} \sup_{|x|<1}\left| f_n(x)-f(x) \right|&=\sup_{|x|<1}\left| f_n(x) \right| \\ &=\sup_{|x|<1} \left| \frac{x^{4n}}{4+x^{4n}} \right| \\ &\to 0 \ \text{as} \ n \to \infty. \end{align}
And the second case would be when $|x|\geq1$:
\begin{align} \sup_{|x|\geq 1}\left| f_n(x)-f(x) \right| &=\sup_{|x|\geq 1} \left| \frac{x^{4n}}{4+x^{4n}} -1 \right| \\ &=\sup_{|x| \geq 1} \left| \frac{-4}{4+x^{4n}} \right| \\ &\to 0 \ \text{as} \ n \to \infty. \end{align}
So to conclude, as $\sup_{x \in \mathbb{R}}\left| f_n(x)-f(x) \right|\to 0 \ \forall x \in \mathbb{R}$, $f_n \to f$ uniformly on $\mathbb{R}$.
Is this correct?
The $f_n$ are continuous functions on $\mathbb{R}$, but the poinwise limit of $f_n(x)$ is not, since the uniform convergence preserve the continuity you can conclude that $f_n$ can't converge uniformly on all $\mathbb{R}$.