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

Observe that if $V(r)=\frac{A}{e^{Br}}-\frac{C}{r^6}$ approaches positive/negative infinity, then either $\frac{A}{e^{Br}}$ or $-\frac{C}{r^6}$ 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 $\frac{A}{e^{Br}}$ or of $-\frac{C}{r^6}$.

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

For horizontal asymptotes, since $V(r)$ is only defined for $r>0$, we find that $$\lim {}_{r\to +\mathrm{\infty }}V(r)=\lim {}_{r\to +\mathrm{\infty }}(\frac{A}{e^{Br}}-\frac{C}{r^6})=0-0=0.$$ So the only horizontal asymptote is $V=0$.

Answer: the vertical asymptote is $r=0$ and the horizontal asymptote is $V=0$.