Evaluate the limit, if it exists. (If an answer does not exist, enter DNE.)

lim (x + h)^3 − x^3 (over) h
h → 0

Question

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

solomonzelman · Answer

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

solomonzelman · Answer

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

anonymous · Answer

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

solomonzelman · Answer

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

solomonzelman · Answer

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

solomonzelman · Answer

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

solomonzelman · Answer

(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.)

solomonzelman · Answer

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