Solving a limit with two variables

Question

let $f(x)=x^3$

How do I solve this limit? $$\lim_{h\to0}\dfrac{f(x+h)-f(x)}{h}$$

I can replace the function with its content

$$\lim_{h\to0}\dfrac{(x+h)^3-x^3}{h}$$

Then expand the paranthesis $$\lim_{h\to0}\dfrac{x^3+3h^2x+3hx^2+h^3-x^3}{h}$$

Thus simplifying to $$\lim_{h\to0}\dfrac{3h^2x+3hx^2+h^3}{h}$$

From here, I don't know where to go in order to solve this limit. I feel like I made the problem more complex than actually simplifying it.

Answer 1

You are almost done indeed note that form here

$$\lim_{h\to0}\dfrac{3h^2x+3hx^2+h^3}{h}=\lim_{h\to0} \,(3hx+3x^2+h^2)=0+3x^2+0=3x^2$$

that is exactly the derivative for $x^3$.

Answer 2

The limit is not in two variables, because $x$ is constant. When you arrive at $$ \lim_{h\to0}\frac{3hx^2+3h^2x+h^3}{h}=\lim_{h\to0}(3x^2+3xh+h^2) $$ you can apply standard rules of limits: \begin{align} \lim_{h\to0}(3x^2+3xh+h^2) &=\lim_{h\to0}3x^2+\lim_{h\to0}3xh+\lim_{h\to0}h^2\\[6px] &=3x^2+3x\lim_{h\to0}h+\Bigl(\,\lim_{h\to0}h\Bigr)^{\!2}\\[6px] \end{align} provided all limits after the top equals sign exist finite. Since they do, you can conclude with $$ =3x^2 $$ Usually this is applied without doing the splitting into “more elementary” limits, because the same rules applied before tell us that polynomials are continuous functions, so their limit at a point can be computed by simple substitution.