How do you find the limit of  (1/(h+2)^2 - 1/4) / h as h approaches 0?

Mar 17, 2018

We need first to manipulate the expression to put it in a more convenient form

Explanation:

Let's work on the expression

$\frac{\frac{1}{h + 2} ^ 2 - \frac{1}{4}}{h} = \frac{\frac{4 - {\left(h + 2\right)}^{2}}{4 {\left(h + 2\right)}^{2}}}{h} = \frac{\frac{4 - \left({h}^{2} + 4 h + 4\right)}{4 {\left(h + 2\right)}^{2}}}{h} = \frac{\frac{\left(4 - {h}^{2} - 4 h - 4\right)}{4 {\left(h + 2\right)}^{2}}}{h} = \frac{- {h}^{2} - 4 h}{4 {\left(h + 2\right)}^{2} h} = \frac{h \left(- h - 4\right)}{4 {\left(h + 2\right)}^{2} h} = \frac{- h - 4}{4 {\left(h + 2\right)}^{2}}$

Taking now limits when $h \to 0$ we have:

${\lim}_{h \to 0} \frac{- h - 4}{4 {\left(h + 2\right)}^{2}} = \frac{- 4}{16} = - \frac{1}{4}$