Given $R\in\mathbb{R}$, $R>1$ investigate the sequence of functions $(f_n)_{n\in\mathbb{N}}$, given by
$$f_n(x)=\frac{x^{2n}}{1+x^{2n}}, \qquad x \in\ [R,\infty)$$
with regard to uniform convergence.
Could someone help me with this problem? I got that the limit of $f_n(x)$ converges to $1$. But does this mean, that $f_n(x)$ is not pointwise convergent and that's why also not uniformly convergent?
Best Answer
The question has been edited.
$f_n(x)$ converges point-wise to $1$. $|f_n(x)-1|=\frac 1 {1+x^{2}n} \leq \frac 1{1+R^{2}n}$ and $\frac 1 {1+R^{2}n} \to 0$. Hence the convergence is uniform.
Answer for the revised version:
$f_n(x)$ converges point-wise to $1$. $|f_n(x)-1|=\frac 1 {1+x^{2n}} \leq \frac 1{1+R^{2n}}$ and $\frac 1 {1+R^{2n}} \to 0$. Hence the convergence is uniform.