Question #b676d

Mar 2, 2015

First of all let's find the domain of the function: the only condition is that the denominator has to be not-zero.

${e}^{x} - 7 \ne 0 \Rightarrow {e}^{x} \ne 7 \Rightarrow x \ne \ln 7$, so the domain is:

$D = \left(- \infty , \ln 7\right) \cup \left(\ln 7 , + \infty\right)$.

Now let's calculate all the limits:

${\lim}_{x \rightarrow - \infty} \frac{2 {e}^{x}}{{e}^{x} - 7} = {0}^{+} / \left({0}^{+} - 7\right) = {0}^{-}$

and this means that $y = 0$ is an horizontal asymptote;

${\lim}_{x \rightarrow {\left(\ln 7\right)}^{\pm}} \frac{2 {e}^{x}}{{e}^{x} - 7} = \frac{2 {e}^{{\left(\ln 7\right)}^{\pm}}}{{e}^{{\left(\ln 7\right)}^{\pm}} - 7} =$

$= \frac{2 \cdot {7}^{\pm}}{{7}^{\pm} - 7} = {14}^{\pm} / {0}^{\pm} = \pm \infty$

and this means that $x = \ln 7$ is a vertical asymptote;

${\lim}_{x \rightarrow + \infty} \frac{2 {e}^{x}}{{e}^{x} - 7} = \frac{+ \infty}{+ \infty - 7} = \frac{+ \infty}{+ \infty} = 2$,

because the two infinites are of the same order, so the limit is the ratio of the two coefficients ($\frac{2}{1}$),

and this means that $y = 2$ is an other horizontal asymptote.

And this is the graph:

graph{2e^x/(e^x-7) [-14.24, 14.23, -7.11, 7.13]}