# How do you evaluate the limit (x^2-9)/(x-3) as x approaches 3?

Dec 26, 2016

$6$

#### Explanation:

Factorise and simplify the function.

${\lim}_{x \to 3} \frac{\left(\cancel{x - 3}\right) \left(x + 3\right)}{\cancel{x - 3}} = {\lim}_{x \to 3} \left(x + 3\right) = 6$

Dec 27, 2016

${\lim}_{x \to 3} \frac{{x}^{2} - 9}{x - 3} = 6$

#### Explanation:

A simpler method is to apply L'Hopitals rule if you get a $\frac{0}{0}$ indeterminate form when evaluating your expression at the limit.

Since ${\lim}_{x \to 1} \frac{{x}^{2} - 9}{x - 3} = \frac{{3}^{3} - 9}{3 - 3} = \frac{0}{0}$ we can apply L'Hopitals Rule.

L'Hopitals rule states the limit of an indeterminate form can be calculated by taking the limit of the derivative of the numerator over the derivative of the denominator like so:

${\lim}_{x \to 3} \frac{{x}^{2} - 9}{x - 3} = {\lim}_{x \to 3} \frac{\left({x}^{2} - 9\right) '}{\left(x - 3\right) '} = {\lim}_{x \to 3} \frac{2 x}{1}$

${\lim}_{x \to 3} \frac{2 x}{1} = \frac{2 \cdot 3}{1} = 6$