Answer
$\lim _{x\rightarrow 0}\left( \dfrac {1}{x}+\dfrac {2}{x^{2}-2x}\right) =-\dfrac {1}{2}$
Work Step by Step
$\lim _{x\rightarrow 0}\left( \dfrac {1}{x}+\dfrac {2}{x^{2}-2x}\right) =\lim _{x\rightarrow 0}\dfrac {x-2+2}{x\left( x-2\right) }=\lim _{x\rightarrow 0}\dfrac {1}{x-2}=\dfrac {1}{0-2}=-\dfrac {1}{2}$
