## Calculus: Early Transcendentals 8th Edition

$\lim\limits_{h \to 0}\frac{(2+h)^5-32}{h}$ We could guess that the value of the limit is $80.0$
$\lim\limits_{h \to 0}\frac{(2+h)^5-32}{h}$ 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 $80.0$