1. ## Another limit question

lim as h -> 0 (6 + h) to the power of 3 subtract 216 divided by h

Any help?

2. You can expand $\displaystyle (6+h)^3$ or take advantage of property $\displaystyle a^3-b^3=(a-b)(a^2+ab+b^2)$ on numerator.

3. Originally Posted by Hockey_Guy14
lim as h -> 0 (6 + h) to the power of 3 subtract 216 divided by h

Any help?
This is apparently
$\displaystyle \lim_{h \to 0} \frac{(6 + h)^3 - 216}{h}$

Expand the cubic:
$\displaystyle = \lim_{h \to 0} \frac{(216 + 108h + 18h^2 + h^3) - 216}{h}$

Now simplify:
$\displaystyle = \lim_{h \to 0} \frac{108h + 18h^2 + h^3}{h}$

Factor:
$\displaystyle = \lim_{h \to 0} \frac{h(108 + 18h + h^2)}{h}$

Cancel:
$\displaystyle = \lim_{h \to 0} (108 + 18h + h^2)$

And finally take the limit:
$\displaystyle = 108$

Many limits in the "secant form" above can be done by this procedure.

-Dan