Answer
$-\frac{1}{25}=-0.04$
Work Step by Step
$\lim _{h\rightarrow 0}\dfrac {\dfrac {1}{5+h}-\dfrac {1}{5}}{h}=\dfrac {\dfrac {5-\left( 5+h\right) }{5\times \left( 5+h\right) }}{h}=\dfrac {-h}{5\times \left( 5+h\right) \times h}=\dfrac {-1}{5\times \left( 5+h\right) }=\dfrac {-1}{5\times (5+0)}=-\frac{1}{25}$
