So we had the pointwise and uniform convergence, and I do get that a sequence of function can converge to a function, just like ordinary sequences do. But what I don't quite get is this pointwise and uniform convergence...
For example, if we take the functions $$f_n,g_n,h_n:\mathbb{R}\rightarrow\mathbb{R} \qquad f_n(x)=\frac{x^{2n}}{1+x^{2n}}, g_n(x)=e^{-n\cdot x^2},h(x)=\sqrt{\frac{1}{n}+x^2}.$$ I have calculated the functions to which they converge pointwise: $$f(x) = \begin{cases} 0 & \textrm{ for $\lvert x\rvert<1$}\\ \frac{1}{2} & \textrm{ for $\lvert x\rvert=1$} \\ 1 & \textrm{ for $\lvert x\rvert>1$}\\ \end{cases}, g(x)=0,h(x)=\lvert x \rvert.$$
The only thing left is to see if they also converge uniform.
For $g_n(x)$ it's not hard to see that for any $\epsilon<\frac{1}{e}$ we have that $$\lvert f_n(\frac{1}{\sqrt{n}})-f(\frac{1}{\sqrt{n}})\rvert=e^{-1}>\epsilon,$$ which should be enough to show that it doesnt converges uniform.
For the other two functions I don't know how to prove or disprove their uniform convergence, but I do have a feeling that they are both uniformly convergent...
Is there a trick that I can use to determine here for determing the uniform convergence? Is there any relation to the norm $\lvert\lvert a_n(x)-a(x)\rvert\rvert$, where $a_n(x)$ converges to $a(x)$ pointwise?