LIMITing factor

$\begin{aligned} L & = \lim_{x \to 3} \frac {x^3-10x+3}{x^2-9} \\ & = \lim_{x \to 3} \frac {x(x^2-9) - x+3}{(x-3)(x+3)} \\ & = \lim_{x \to 3} x - \frac 1{x+3} \\ & = 3 - \frac 16 = \boxed {\frac {17}6} \end{aligned}$

• We first substitute $x=3$

$\large{\dfrac{x^3-10x+3}{x^2-9}= \dfrac{3^3-10\cdot3+3}{3^2-9}=\dfrac{0}{0}}$

• We get a $\dfrac{0}{0}$ situation, so we can apply L'Hôpital's rule.

• We differentiate both numerator and denominator to get:

$\large{\lim \limits_{x\rightarrow 3} \dfrac{x^3-10x+3}{x^2-9}} = \dfrac{\frac{d}{dx} (x^3-10x+3)}{\frac{d}{dx} (x^2-9)}=\dfrac{3x^2-10}{2x}$

• Now we can substitute $x=3$ and get the answer.

$\large{\lim \limits_{x\rightarrow 3} \dfrac{x^3-10x+3}{x^2-9}} = \dfrac{3\cdot 3^2-10}{2\cdot 3}=\boxed{\dfrac{17}{6}}$