Let $A=[-1,1]$, and $f:A\to\Bbb{R}$ a continuous function, such that $f(0)=0$. I need to show that $g_n$ converges uniformly on $A$, when $$g_n(x) = \frac{f(x)}{1+nx^2}$$
I honestly don't get this one. The only thing that I can think off is pretty far fetched : Supposing the limit function is $f(x) = 0$. Then, because $f(x)$ is continuous, it attains a maximum value on $A$, and therefore so does $g_n(x)$. It seems like $g_n(x)$ is maximized when the denominator is minimized, and the numerator maximized. At this point we might look at what happens when $x\to 0$, apply definitions of continuity etc. The idea is to of course show that the sequence $$M_n=\sup_{x\in A}\vert g_n(x)-0\vert$$
goes to zero as $n\to \infty$. But since I can't deduce what the limit function is, this test will not work. Help please.
The idea is to use $f(0) = 0$ to get a uniform bound of the $g_n$ near the origin, and the general boundedness of $f$ to get a uniform bound on the rest of $A$, away from the origin.
So let $\epsilon > 0$.
First, from the continuity of $f$ and $f(0) = 0$ it follows that there is a $\delta > 0$ such that $|f(x)| < \epsilon$ for $|x| < \delta$. It follows that for $|x| < \delta$ and all $n$ $$ |g_n(x)| \le \frac{\epsilon}{1+nx^2} \le \epsilon \, . $$
Second, $f$ is bounded on $A$, say $|f(x)| \le M$. It follows that for $|x| \ge \delta$ $$ |g_n(x)| \le \frac{M}{1+nx^2} \le \frac{M}{1+n \delta^2} \, , $$ and that is also $< \epsilon$ for sufficiently large $n$.
This shows that $|g_n(x)| \le \epsilon$ for all $x \in A$ and all sufficiently large $n$, i.e. $g_n$ converges uniformly to zero.