When finding the derivative using its definition, why do we plug $0$ for $h$?

Question

If $\lim h\to 0$, when finding the derivative of the function, why do you plug in the limit that is being approached. Like why would you plug in $0$ in the function $4x+2h$ (which is the derivative of $\frac{2 (x+h)^2-2x^2}{h}$

Answer 1

Ok, so I assume that you are trying to find the derivative of function $f(x)=2x^2$. So, as you suggest, we would like to look at the following limit $$\lim_{h\to 0}\frac{2(x+h)^2-2x^2}{h}$$ Expending, we have $$\lim_{h\to 0}\frac{2(x+h)^2-2x^2}{h}= \lim_{h\to 0}(4x+2h)=4x+\lim_{h\to 0} 2h=4x$$

Here, you are NOT taking $h=0$ but rather computing what is the value of $h$ as $h$ goes to 0.

Answer 2

While the other answers are sufficient to tell you how to evaluate the limit, I think you're not sure why we take that limit as $h\to\color{blue}{0}$. So $\Delta x$ is your $h$. What you're looking for is intuition and nothing can give you better intuition than looking at this pic.

Answer 3

In the answer above given by @wisher, note that the expression is simplified and the factor $h$ is divided out of numerator and denominator. Once that is done, what remains is a simple polynomial. In that case, you really can just plug in $h=0$ to get the limit.

Why? Because polynomials are continuous, so for a polynomial $p$, it is true that $$ \lim_{ h\to a} p(h) =p(a).$$