When to use $\lim_{x\to a}\frac{f(x)-f(a)}{x-a}$ vs $\lim_{h\to0}\frac{f(a+h)-f(a)}{h}$ to find the slope of the tangent line I was given 2 formulas but I am unsure of when to use each one. Both of them got the same answer for a given question

Find the slope of tangent at $x = 2$ for the function $y = 2x^2$

Formula 1:
$$\lim_{x \to a} \frac{f(x)-f(a)}{x - a}$$
Formula 2:
$$\lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$$
Answer 1:
$$\lim_{x \to 2} f(x) = 8$$
Answer: 2
$$\lim_{h \to 0} f(x) = 8$$
My question is:

What is the difference between the two answers/formulas? And when should I be using each method?

 A: They are the same formula, except for a shift in the limiting variable. Specifically, if you let $x = a+h$, then both are the same, since $$\lim_{x \to a} \frac{f(x)-f(a)}{x-a} = \lim_{a+h \to a} \frac{f(a+h)-f(a)}{(a+h)-a} = \lim_{h \to 0} \frac{f(a+h)-f(a)}{h}$$
I prefer to use the second formula (with $h$) almost always. I find it easier to find a general derivative as well using this one, but it is up to your personal preference about which one to use.
A: I always found this example enlightening. Let $f(x)=\sin(x)$, then Formula 1 gives
$$f'(a)=\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=\lim_{x\to a}\frac{\sin(x)-\sin(a)}{x-a}.$$
First thoughts are you could try Taylor series, which seems very messy. I also thought about perhaps "reverse engineering" $$\sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x)\tag{1}$$ but that doesn't seem obvious either.
But if we try Formula 2,
$$f'(x)=\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{\sin(x+h)-\sin(x)}{h}=\lim_{h\to 0}\frac{\sin(h)\cos(x)+\sin(x)\cos(h)-\sin(x)}{h},$$
then things look a bit more straightforward since we cam stick to working with limits and show that $\sin(h)/h\to 1$ and $(\sin(x)\cos(h)-\sin(x))/h\to 0$.
As the previous answer says, they are equivalent and one version may be easier to work with than the other, depending on preference.
