Pointwise convergence V.S. Uniform convergence 
My teaching assistant explained that for $f_k$, depicted like the above picture, $f_k$ pointwisely converges like the picture, but $f_k$ does not uniformly converge.

I could guess $f_k(x)=1-kx$.
Then, $\displaystyle\lim_{k\to\infty}f_k(x)=y\mbox{-axis}$.

How can I approach the case like the above?
Could someone make me understand the teaching assistant's picture?
Thank you for reading my question.
 A: I think the functions are $$f_n(x) = \begin{cases}1-nx, &0 < x \leq \frac{1}{n} \\ 0, & x > \frac{1}{n} \end{cases}$$
Let $x_0 \in \mathbb R^+$. Then clearly, $\lim_{n \to \infty} f_n(x_0) =0$.
So, the pointwise limit is $f(x)=0$
However, the graphs of the functions $f_n(x)$ and their limit $f(x)$ don't get close for every $x$.
That is, whatever $f_n(x)$ you draw, that graph will still have points far away from the graph of $f(x)$.
For instance, take $n=10$. Draw the graph of $f_{10}(x)$. Is every point of that function close to the limit function $f(x)$? No.
For example, $f_{10} \left( \frac{1}{20} \right) = \frac 12$. However $f \left( \frac{1}{20} \right) =0$.
So, since $\frac{1}{2}$ and $0$ are "not close", the convergence is not uniform.
By the way, if you add $0$ to the domain of $f_n(x)$, then the limit function will be
$$f(x) = \begin{cases} 1, &x=0 \\ 0, &x >0 \end{cases}$$
But there is a theorem saying that "unform limit of continuous functions is continuous", so this convergence cannot be uniform.
A: Another characterization which will help you a lot...let $M_n=\sup_x|f_n(x)-f(x)|$. Then $f_n\rightarrow f$ uniformly on the given set if   $M_n\rightarrow 0$. In your question $M_n=1$ ..:)
