Science:Math Exam Resources/Courses/MATH 180/December 2017/Question 06/Solution 1

Observe that if $V(r)={\dfrac {A}{e^{Br}}}-{\dfrac {C}{r^{6}}}$ approaches positive/negative infinity, then either ${\dfrac {A}{e^{Br}}}$ or $-{\dfrac {C}{r^{6}}}$ approaches positive/negative infinity, as $r\to a$, for some $a$. In other words, if $r=a$ is a vertical asymptote of $V$, then $r=a$ is a vertical asymptote of ${\dfrac {A}{e^{Br}}}$ or of $-{\dfrac {C}{r^{6}}}$.

Recall that the exponential function can never be zero, so ${\dfrac {A}{e^{Br}}}$ is finite for any $r$. On the other hand, the rational function $-{\dfrac {C}{r^{6}}}$ is not defined at $r=0$, which makes $r=0$ a candidate for a vertical asymptote of $-{\dfrac {C}{r^{6}}}$ and hence of $V$. In fact,

$$\lim \limits _{r\to 0^{+}}V(r)=\lim \limits _{r\to 0^{+}}\left({\dfrac {A}{e^{Br}}}-{\dfrac {C}{r^{6}}}\right)=A-\lim \limits _{r\to 0^{+}}{\dfrac {C}{r^{6}}}=-\infty .$$

So ${\textstyle r=0}$ is the only vertical asymptote.

For horizontal asymptotes, since ${\textstyle V(r)}$ is only defined for ${\textstyle r>0}$, we find that

$$\lim \limits _{r\to +\infty }V(r)=\lim \limits _{r\to +\infty }\left({\dfrac {A}{e^{Br}}}-{\dfrac {C}{r^{6}}}\right)=0-0=0.$$

So the only horizontal asymptote is ${\textstyle V=0}$.

Answer: the vertical asymptote is ${\textstyle {\color {blue}r=0}}$ and the horizontal asymptote is ${\textstyle {\color {blue}V=0}}$.