# Pointwise vs Uniform Convergence

Let {$$f_n$$ } be a sequence of functions defined by $$f (x) =\frac{x}{1+n^2x^2}$$ on $$[0, 1]$$. Prove that

1. {$$f_n$$} converges uniformly on $$[0,1]$$ to a function $$f$$ that is differentiable on $$[0,1]$$
2. {$$f′_n$$ }converges pointwise on $$[0,1]$$ to a function $$g$$ that is not equal to $$f′$$.

I have proved that $$f_n$$ converges uniformly to $$f(x) = 0$$.

However, I've realized that {$$f′_n$$ } = $$\frac{1 - n^2x^2}{(1+n^2x^2)^2}$$ converges uniformly to $$f(x) = 0$$, and hence converges pointwise to $$f(x) = 0$$, which is equal to $$f'$$. Am I doing something wrong?

The derivatives do not converge uniformly to $$0$$. Instead, they converge pointwise to $$g(x) = \left\{ \array{0 &\text{ if } x\in (0,1] \\ 1 & \text{ if } x=0 }\right.$$
(Note that $$f^\prime_n(0) = 1$$ for every $$n$$).