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