Here is the limit:
$\operatorname*{lim}_{x\rightarrow2}\frac{x^{2}-4}{x-2}$
I know I should match the limit to the definition, but I do not know how to match this to
$\frac{d}{{dx}}f\left( x \right) = \mathop {\lim }\limits_{\Delta \to 0} \frac{{f\left( {x + \Delta } \right) - f\left( x \right)}}{\Delta }$
As noted by the comment $f′(x)=\frac{f(x)-f(x_0)}{x-x_0}$ so we have $x_0=2$ and $f(x)-f(x_0)=x^2-4$ so $f(x)=x^2-4+f(x_0) $ we can actually set $f(x_0)$ to an arbitrary value since shifting the function up and down shouldn't change the derivative at that point so some natural values to choose would be $f(x_0)=0$ or maybe $f(x_0)=4$ which would give us either:$f(x)=x^2-4$ or $f(x)=x^2$ but really anything of the form $f(x)=x^2-4+f(x_0)$ where $x_0=2$ would be the solution to your question.