## anonymous one year ago Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.) lim (x + h)^3 − x^3 (over) h h → 0

1. SolomonZelman

$$\large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{(x+h)^3-x^3}{h}}$$ (Or the derivative of x$$^3$$, for which you should get 3x$$^2$$, by the power rule, thus we know what value must your limit be equal to. --> If you have ever learned the power rule)

2. SolomonZelman

You have to expand the $$(x+h)^3$$, at first.

3. anonymous

$h^3+3h^2x+3hx^2+x^3$

4. SolomonZelman

Yes, and now, write that on top of your fraction instead of (x+h)³.

5. SolomonZelman

$$\large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{(x+h)^3-x^3}{h}}$$ $$\large\color{slate}{\displaystyle\lim_{h \rightarrow ~0}\frac{h^3+3h^2x+3hx^2+x^3-x^3}{h}}$$

6. SolomonZelman

$$x^3$$ goes away, and then h will cancel.

7. SolomonZelman

(The validity of dividing by h on top and bottom, as h$$\rightarrow$$0, is justified by the fact that you are taking values that are not actually equal to 0, rather near 0.)

8. SolomonZelman

$$\large\color{black}{\displaystyle\lim_{h \rightarrow ~0}\frac{h^3+3h^2x+3hx^2\cancel{~+x^3-x^3~}}{h}}$$ $$\large\color{black}{\displaystyle\lim_{h \rightarrow ~0}\frac{h^3+3h^2x+3hx^2}{h}}$$