If you have a function $f$, differentiable in $\mathbb{R}$, such that $y= -3x+2$ is tangent to its graph at $x=2$.
Can you replace $f(x)$ by $-3x+2$ when you are calculating the following limit?
$\lim\limits_{x \to 2} \dfrac{f(x)+4}{x^2-4}$
My opinion is: no, because they are not the same thing. But it works and I can't find an example where it doesn't work. Anyone has one?
By the way, what I find formally correct is something like this:
$\lim\limits_{x \to 2} \dfrac{f(x)+4}{x^2-4} = \underbrace{\lim\limits_{x \to 2} \dfrac{f(x)-(\color{red}{\overbrace{\color{black}{-4}}^{f(2)}})}{x-2}}_{f'(2)} \times \lim\limits_{x \to 2} \dfrac{1}{x+2} = -3\times \dfrac{1}{4} = -\dfrac{3}{4}$
You are right, this replacement would not be rigourous. What you can do instead is apply the definition of $f'(2)$ you recalled, and which can be reworded as $$\frac{f(x)-f(2)}{x-2}=f'(2)+\varepsilon(x)\text{ whith }\lim_{x\to2}\varepsilon(x)=0,$$ i.e. replace $f(x)$ with $$f(2)+(x-2)f'(2)+(x-2)\varepsilon(x)=-3x+2+(x-2)\varepsilon(x).$$ This way, as $x\to2,$ $$\frac{f(x)+4}{x^2-4}=\frac{-3x+6+(x-2)\varepsilon(x)}{x^2-4}=\frac{-3+\varepsilon(x)}{x+2}\to-\frac34.$$
This is more than a matter of rigor. A rough replacement of $f(x)$ by its linear approximation can lead to wrong results. Take for instance $f(x)=\cos x$ and look at $\lim_{x\to0}\frac{f(x)-1}{x^2}.$ If you replace $f(x)$ with $f(0)+xf'(0)=1+0x=1,$ the numerator becomes $0$ and you find a limit equal to $0,$ whereas the correct result is $-\frac12.$