Answer
$-\dfrac {8}{225} $
Work Step by Step
$\lim _{x\rightarrow 3}\dfrac {\dfrac {1}{x^{2}+2x}-\dfrac {1}{15}}{x-3}=\dfrac {15-x^{2}-2x}{15\left( x^{2}+2x\right) \times \left( x-3\right) }=\dfrac {9+6-x^{2}-2x}{15\left( x^{2}+2x\right) \left( x-3\right) }=\dfrac {3^{2}+6-x^{2}-2x}{15\left( x^{2}+2x\right) \left( x-3\right) }=\dfrac {\left( 3-x\right) \left( 3+x\right) +\left( 6-2x\right) }{15\left( x^{2}+2x\right) \left( x-3\right) } =\dfrac {\left( 3-x\right) \left( 3+x\right) +2\left( 3-x\right) }{15\left( x-3\right) \left( x^{2}+2x\right) }=\dfrac {\left( 3-x\right) \left( 3+x+2\right) }{15\left( x-3\right) \left( x^{2}+2x\right) }=\dfrac {-\left( 5+x\right) }{15\left( x^{2}+2x\right) }=\dfrac {-\left( 5+3\right) }{15\left( 3^{2}+6\right) }=\dfrac {-8}{225} $
