Answer
$\lim\limits_{h \to 0}\frac{(2+h)^5-32}{h}$
We could guess that the value of the limit is $80.0$
Work Step by Step
$\lim\limits_{h \to 0}\frac{(2+h)^5-32}{h}$
We can evaluate the function at the given numbers:
$h = 0.5$:
$\frac{(2+0.5)^5-32}{0.5} = 131.3125$
$h = -0.5$:
$\frac{(2+h)^5-32}{h} = 48.8125$
$h = 0.1$:
$\frac{(2+0.1)^5-32}{0.1} = 88.4101$
$h = -0.1$:
$\frac{(2+h)^5-32}{h} = 72.3901$
$h = 0.01$:
$\frac{(2+h)^5-32}{h} = 80.804010$
$h = -0.01$:
$\frac{(2+h)^5-32}{h} = 79.203990$
$h = 0.001$:
$\frac{(2+h)^5-32}{h} = 80.080040$
$h = -0.001$:
$\frac{(2+h)^5-32}{h} = 79.920040$
$h = 0.0001$:
$\frac{(2+h)^5-32}{h} = 80.008000$
$h = -0.0001$:
$\frac{(2+h)^5-32}{h} = 79.992000$
We could guess that the value of the limit is $80.0$
