# How do you find the limit lim_(h->0)((4+h)^2-16)/h ?

Sep 20, 2014

${\lim}_{h \to 0} \frac{{\left(4 + h\right)}^{2} - 16}{h} = 8$

Let us look at some details.

${\lim}_{h \to 0} \frac{{\left(4 + h\right)}^{2} - 16}{h}$

by multiplying out the numerator,

$= {\lim}_{h \to 0} \frac{16 + 8 h + {h}^{2} - 16}{h}$

by cancelling out $16$'s,

$= {\lim}_{h \to 0} \frac{8 h + {h}^{2}}{h}$

by factoring out $h$ from the numerator,

$= {\lim}_{h \to 0} \frac{h \left(8 + h\right)}{h}$

by cancelling out $h$'s,

$= {\lim}_{h \to 0} \left(8 + h\right) = 8 + 0 = 8$