#### Answer

$\delta=\dfrac{1}{402}$

#### Work Step by Step

We see from the graph that in order for $f(x)$ to be within $\epsilon=0.01$ of $L=2$, we must have
$$\dfrac{1}{2.01} \lt x \lt \dfrac{1}{1.99}.$$
Subtracting $c=\dfrac{1}{2}$ from all three sides gives
$$-\dfrac{0.01}{4.02} \lt x-\dfrac{1}{2} \lt \dfrac{0.01}{3.98}.$$
$$-\dfrac{1}{402} \lt x-\dfrac{1}{2} \lt \dfrac{1}{398}.$$
Note that
$$-\dfrac{1}{402} \lt x-\dfrac{1}{2} \lt \dfrac{1}{402} \implies -\dfrac{1}{402} \lt x-\dfrac{1}{2} \lt \dfrac{1}{398}.$$
Hence for $\delta = \dfrac{1}{402}$,
$$0 \lt \left|x-\dfrac{1}{2}\right| \lt \delta \implies 0 \lt |f(x)-2| \lt \epsilon.$$