No. For x=2 you have:
[tex]f(2)=\dfrac{2^2-4}{2-2}=\dfrac{4-4}{2-2}=\dfrac{0}{0}[/tex]
This is the indeterminate form 0/0 and its value could be anything you want (0, any number or infinity).
But:
[tex]\lim_{x\to2}\dfrac{x^2-4}{x-2}=\lim_{x\to2}\dfrac{(x+2)(x-2)}{x-2}=\lim_{x\to2}(x+2)=4[/tex]
So answer is B.
You could see that when you calculate
f(1.999) ≈ 3.999
and
f(2.001) ≈ 4.001
