Weak convergence of a sequence of linear functionals $l_n(f) = n\int_0^\frac{1}{n} f(x)\ \text{d}x$. I'm doing exercise 7.4 of Berezanksky's Functional Analysis Vol I. I need to compute the weak limit, $l$, of a sequence of linear functionals $(l_n)_{n=1}^\infty$ defined on $C[0,1]$ with the sup-norm where
$$
l_n(f) = n\int_0^\frac{1}{n} f(x)\ \text{d}x 
$$
I have little intuition for what the weak limit of this is. I first thought that $l(f)=0$ for all $f$, then $l(f)=\|f\|_\infty$. I couldn't prove either of these. I've been told that $l(f)=f(0)$. This makes sense but I'm having trouble showing it. Here's my attempt:
\begin{align*}
|l_n(f)-l(f)| &= \bigg|n\int_0^\frac{1}{n} f(x)\ \text{d}x-f(0)\bigg| \\
&= \bigg|n\int_0^\frac{1}{n} f(x)\ \text{d}x - n\int_0^\frac{1}{n} f(0)\ \text{d}x\bigg| \\
&= \bigg|n\int_0^\frac{1}{n} f(x)-f(0)\ \text{d}x\bigg| \\
&\leq n\int_0^\frac{1}{n}|f(x)-f(0)|\ \text{d}x
\end{align*}
Now it feels like I need to make some observation about the final line to see that $|l_n(f)-l(f)|$ can be made arbitrarily small. I've drawn some diagrams and cannot see this. Is my direction even correct?

Hint: What does the continuity of $f$ mean at $x=0$?  
Another way: From the MVT, we have that
$$\int_0^{1/n} f=\frac{1}{n}f(c)$$
for some $c \in (0, 1/n)$. This means that
$$n\int_0^{1/n} f=f(c)$$
Which gives us that
$$n\int_0^{1/n} f \to f(0)$$
as desired.
