Both parts of this are actually quite tricky.

Good start. So we now want to show that if 0<x<1 then

as

. As you can see, this is not obvious. One sneaky way to prove it is to notice that the series

converges if |x|<1 (easily shown by using the ratio test), and therefore the n'th term of the series must tend to 0.

To say that a sequence of functions {f_n(x)} converges uniformly to zero is the same as saying that the maximum value of |f_n(x)| tends to 0 as n→∞. In this case, the functions are all non-negative, so we can drop the absolute value signs and ask whether the maximum of f_n(x) goes to 0. By basic calculus, the max value of

in the unit interval occurs when the derivative is 0. This occurs when

, at which point the value of the function is

, which converges to 1/e as n→∞. Since this is greater than 0, the sequence of functions does not converge uniformly.

As I said, it's a tricky question.