Left-hand derivative at x = 2

For $x$ near $2$ from the left, $f(x) = x^2$, so $f(2) = 4$. The difference quotient is $\frac{x^2 - 4}{x - 2} = \frac{(x-2)(x+2)}{x-2} = x + 2,$ valid for $x \ne 2$.

Taking the limit as $x \to 2^-$ gives $x + 2 \to 4$. So the left-hand derivative at $x=2$ is $4$.

For $x > 2$, $f(x) = 4x - 4$, and $f(2) = 4$ from the definition. The difference quotient is $\frac{(4x-4) - 4}{x-2} = \frac{4x - 8}{x - 2} = \frac{4(x-2)}{x-2} = 4,$ valid for $x \ne 2$.

This expression equals $4$ for all $x$ near $2$ (not just in the limit), so the limit as $x \to 2^+$ is simply $4$.

From part (a), the left-hand limit of the difference quotient is $4$, and from part (b), the right-hand limit is also $4$.

Since both one-sided limits agree and equal $4$, the two-sided limit $\lim_{x\to 2}\frac{f(x)-f(2)}{x-2}$ exists and equals $4$. Therefore $f$ is differentiable at $x=2$ with $f'(2) = 4$, so $f$ has a tangent line there with slope $4$.