Question

lim as h->0 to ((5+h)^-1 - 5^-1)/h?

Answer 1

if you want to make sense out of this get rid of the stupid negative exponents.
\[\lim_{h \rightarrow 0}\frac{\frac{1}{5+h}-\frac{1}{5}}{h}\]

Answer 2

then do the arithmetic in the numerator
\[\frac{1}{5+h}-\frac{1}{5}=\frac{5-(5+h)}{5(5+h)}=\frac{-h}{5(5+h)}\]\]

Answer 3

divide by h ( they cancel) to get
\[\frac{-1}{5(5+h)}\] then take the limit by replacing h by 0 to get
\[\lim_{h\rightarrow 0}\frac{-1}{5(5+h)}=-\frac{1}{25}\]