Question

can someone please help me solve this limit

lim [1/(8+x) -1/8]/x
X-->0

Answer 1

\[\lim_{x \rightarrow 0} \frac{\frac{1}{8+x}-\frac{1}{8}}{x}\] It is a \[\frac{0}{0}\]
You probably should apply L'Hopital
\[\lim_{x \rightarrow 0} \frac{\frac{1}{-(8+x)^2}-0}{1}=\lim_{x \rightarrow 0} -\frac{1}{(8+x)^2}=\frac{1}{64}\]

Answer 2

have never seen the law , i was trying to get common denom and cancel out but got stuck, thanks still