# For what values of x, if any, does f(x) = 1/((x-3)(x-7))  have vertical asymptotes?

Jan 7, 2016

$x = 3$ vertical asymptote
$x = 7$ vertical asymptote

graph{1/((x-3)(x-7)) [-10, 10, -5, 5]}

#### Explanation:

Vertical asymptotes could be find in the points where the Exist Conditions of $f \left(x\right)$ are not satisfied

The Field of Existence of $f \left(x\right)$ is:

x in ]-oo,3[uu]3,7[uu]7,+oo[

because

$f \left(x\right) = \frac{N \left(x\right)}{D \left(x\right)}$

FE (Field of Existence):

$D \left(x\right) \ne 0$

$\left(x - 3\right) \left(x - 7\right) \ne 0$

${x}_{1} \ne 3$
${x}_{2} \ne 7$

Now these values of $x$ are vertical asymptotes if:

${\lim}_{x \rightarrow {x}_{1 , 2}^{\pm}} f \left(x\right) = \pm \infty$

${\lim}_{x \rightarrow {3}^{-}} f \left(x\right) = {\lim}_{x \rightarrow {3}^{-}} \frac{1}{\left(2.9 - 3\right) \left(2.9 - 7\right)} =$
${\lim}_{x \rightarrow {3}^{-}} \frac{1}{\left({0}^{-}\right) \left(- 4.1\right)} = {\lim}_{x \rightarrow {3}^{-}} \frac{1}{0} ^ + = + \infty$

${\lim}_{x \rightarrow {3}^{+}} f \left(x\right) = {\lim}_{x \rightarrow {3}^{+}} \frac{1}{\left(3.1 - 3\right) \left(3.1 - 7\right)} =$
${\lim}_{x \rightarrow {3}^{+}} \frac{1}{\left({0}^{+}\right) \left(- 3.9\right)} = {\lim}_{x \rightarrow {3}^{+}} \frac{1}{{0}^{-}} = - \infty$

${\lim}_{x \rightarrow {7}^{-}} f \left(x\right) = {\lim}_{x \rightarrow {7}^{-}} \frac{1}{\left(6.9 - 3\right) \left(6.9 - 7\right)} =$
${\lim}_{x \rightarrow {7}^{-}} \frac{1}{\left(3.9\right) \left({0}^{-}\right)} = {\lim}_{x \rightarrow {7}^{-}} \frac{1}{{0}^{-}} = - \infty$

${\lim}_{x \rightarrow {7}^{+}} f \left(x\right) = {\lim}_{x \rightarrow {7}^{+}} \frac{1}{\left(7.1 - 3\right) \left(7.1 - 7\right)} =$
${\lim}_{x \rightarrow {7}^{+}} \frac{1}{\left(4.1\right) \left({0}^{+}\right)} = {\lim}_{x \rightarrow {7}^{+}} \frac{1}{{0}^{+}} = + \infty$