Show that a sequence converges in the $d_\infty$ metric. Recall the $d_\infty$ metric on the set of continuous functions $C([0,1],R)$:
$d_\infty(f,g) = \max\{|f(x)−g(x)|:x\in[0,1]\}$
Consider the sequence in $C([0,1],R)$ defined by 
$$f_n(x) = \frac{1}{(n^2)(1+x^{2n})},\quad n \ge 1,\ x \in[0,1]$$
Show that $(f_n)_{n≥1}$ converges in $d_\infty$ metric and find the limit.
 A: The generally accepted way to do this would be to prove that $(f_n)_{1\le n}$ is a Cauchy sequence in the metric space $\langle [0,1],d_\infty\rangle$. That is, for every real number $\epsilon>0$ there exists a natural number $N$ such that for all $m,n>N$, $d_\infty(f_m,f_n)<\epsilon$.
Since you have this tagged as topolical groups, I should also mention there is an extension of the notion of "Cauchy sequence" for topological groups, stated on Wikipedia:

A sequence $(x_k)$ in a topological group $G$ is a Cauchy sequence if
  for every open neighbourhood $U$ of the identity in $G$ there exists
  some number $N $such that whenever $m,n>N$  it follows that $x_n x_m^{-1} \in U$. As above, it is sufficient to check this for the
  neighbourhoods in any local base of the identity in $G$.

I don't know anything about topological groups, and I've only every used the open set definition of topology, so I'm going to stick with the metric space definition.
Thm: The sequence $(f_n)_{1\le n}$ is a Cauchy sequence in the $d_\infty$ metric.
Proof: By definition... $$f_n(x)=\frac{1}{(n^2)(1+x^{2n})}$$
Notice that for any $n\ge 1$, $\max\{f_n(x):x\in[0,1]\}=f_n(0)$, hence...
$$\max\{f_n(x): x\in[0,1]\}=\frac{1}{n^2}$$
Because $\left(\frac{1}{n^2}\right)_{n\in\Bbb{N}}$ is Cauchy, it follows that $(f_n)_{1\le n}$ is Cauchy.

As for the limit, it is clearly $0$.
