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

Nov 10, 2016

$= 6$

#### Explanation:

Substituting $3$ in the given expression, to find its limit, results to
$\text{ }$
an indeterminate solution :

${\lim}_{x \to 3} \frac{\left\mid {x}^{2} - 9 \right\mid}{\left\mid x - 3 \right\mid}$

$= \frac{\left\mid {3}^{2} - 9 \right\mid}{\left\mid 3 - 3 \right\mid}$

$= \frac{0}{0}$ $\text{ }$ Indeterminate solution.
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Factorizing $\text{ "x^2-9" }$ is the ideal way to find the solution.

${x}^{2} - 9 = {x}^{2} - {3}^{2} = \left(x - 3\right) \left(x + 3\right)$
$\text{ }$
$\text{ }$

${\lim}_{x \to {3}^{-}} \frac{\left\mid {x}^{2} - 9 \right\mid}{\left\mid x - 3 \right\mid} = {\lim}_{x \to {3}^{-}} \frac{\left\mid \left(x - 3\right) \left(x + 3\right) \right\mid}{\left\mid x - 3 \right\mid}$
$\text{ }$

=lim_(x->3^-)(abs(x-3)abs(x+3))/abs(x-3)" " =lim_(x->3^-)((3-x)(x+3))/(3-x)
$\text{ }$
$= {\lim}_{x \to {3}^{-}} \frac{\cancel{\left(3 - x\right)} \left(x + 3\right)}{\cancel{\left(3 - x\right)}} = {\lim}_{x \to {3}^{-}} 3 + x = 3 + 3 = 6$
$\text{ }$
$\text{ }$
$\text{ }$

${\lim}_{x \to {3}^{+}} \frac{\left\mid {x}^{2} - 9 \right\mid}{\left\mid x - 3 \right\mid} = {\lim}_{x \to {3}^{+}} \frac{\left\mid \left(x - 3\right) \left(x + 3\right) \right\mid}{\left\mid x - 3 \right\mid}$
$\text{ }$

=lim_(x->3^+)(abs(x-3)abs(x+3))/abs(x-3)" " =lim_(x->3^+)((x-3)(x+3))/(x-3)
$\text{ }$
$= {\lim}_{x \to {3}^{+}} \frac{\cancel{\left(x - 3\right)} \left(x + 3\right)}{\cancel{\left(x - 3\right)}} = {\lim}_{x \to {3}^{+}} x + 3 = 3 + 3 = 6$
$\text{ }$
$\text{ }$
Therefore,
$\text{ }$

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