The function sequence $(f_n)$ with $f_n = \frac{x^{2n}}{1+x^{2n}}$, where $n \in \Bbb{N}$ has three limit functions depending on $x$
$$f(x)=\begin{cases} 0 & \vert x\vert <1 \\ 1/2 & \vert x \vert =1 \\ 1 & \vert x \vert >1 \end{cases}$$
Let $\epsilon >0$. When $\vert x \vert=1$, the sequence remains constant and so even the choice $N=1$ is enough to guarantee that $n\geq N$ implies $\vert f_n(x) -f \vert < \epsilon$. However when $\vert x \vert >1$, the choice $N=1$ is not sufficient. This is because the relation $$\epsilon>\frac{x^2}{1+x^2}-1=\frac{1}{1+x^2}$$
does not hold in general when $\vert x \vert >1$. This shows the sequence will not converge uniformly on $\Bbb{R}$, correct?