# How do you find the lim_(xrarr3) (x^2+x-12)/(x-3)?

Apr 16, 2017

$7$

#### Explanation:

Rewrite the numerator as $\left(x + 4\right) \left(x - 3\right)$

${\lim}_{x \to 3} \frac{\left(x + 4\right) \left(x - 3\right)}{x - 3}$

Cancel the $x - 3$

${\lim}_{x \to 3} x + 4$

Plug in $3$ for $x$:

$3 + 4 = 7$

Apr 16, 2017

The numerator factors and cancels the denominator, leaving a simple linear expression that can be evaluated at the limit.

#### Explanation:

Given: ${\lim}_{x \rightarrow 3} \frac{{x}^{2} + x - 12}{x - 3}$

Factor the numerator:

${\lim}_{x \rightarrow 3} \frac{\left(x - 3\right) \left(x + 4\right)}{x - 3}$

Please observe the factors that cancel:

${\lim}_{x \rightarrow 3} \frac{\cancel{x - 3} \left(x + 4\right)}{\cancel{x - 3}}$

This leaves a simple linear expression that can be evaluated at the limit:

${\lim}_{x \rightarrow 3} x + 4 = 7$

Apr 16, 2017

$7$

#### Explanation:

$\text{factorise numerator and simplify}$

${\lim}_{x \to 3} \frac{\left(x + 4\right) \left(\cancel{x - 3}\right)}{\cancel{x - 3}}$

$= {\lim}_{x \to 3} \left(x + 4\right) = 3 + 4 = 7$