I'm trying to find the asymptotes of

$$f(x) = \exp\left(\frac{3+4x^2}{2-6x}\right).$$

The vertical asymptote was a piece of cake and it's $\frac{1}{3}$.

The horizontal asymptotes is calculated as $x\to\infty$. I know the limit will be $0$, I just can't seem to be able to calculate it.

I tried dividing the numerator and denominator by $x^2$ but that doesn't help me. I got a zero in the denominator or after simplifying the fraction I got $\frac{2}{3}$.

I must be doing something wrong.

  • $\begingroup$ Antonio, why did you edit it like that? Now it doesn't make any sense. $\endgroup$ Commented Oct 9, 2015 at 19:42
  • 1
    $\begingroup$ You may have an easier time of it if you use L'Hopital's Rule on the expression in the exponent. Then you have the form $e^{-\infty}$ as $x\to+\infty$. $\endgroup$ Commented Oct 9, 2015 at 19:46
  • $\begingroup$ My professor doesn't let us use L'Hopital's rule unfortunately. Thanks for the hint. $\endgroup$ Commented Oct 9, 2015 at 19:49
  • 1
    $\begingroup$ It may be simpler to see if you divide top and bottom of the exponent by $x$. Then at the bottom we have something that approaches $-6$, and at the top we have something that blows up, so the exponent becomes very large negative. $\endgroup$ Commented Oct 9, 2015 at 19:50
  • 1
    $\begingroup$ @DanielWaleniak: a positive number (greater than $1$) with an exponent tending toward negative infinity will go to $0$. $\endgroup$
    – Clayton
    Commented Oct 9, 2015 at 20:02

1 Answer 1


Hints: what is the first order approximation of $3+4x^2$ towards infinity ? what about $2-6x$ ?

Formally: factor by highest degree term on numerator and denominator. then evaluate the behavior at infinity.

  • $\begingroup$ I got $\frac {4}{0}$ in the exponent when I substituted x for infinity. $\endgroup$ Commented Oct 9, 2015 at 19:58
  • $\begingroup$ let x in place. $\endgroup$ Commented Oct 9, 2015 at 19:59
  • $\begingroup$ Would it be something like $\frac {0+4}{0-6x}$ in the exponent giving $ \frac{-2x}{3}$ which would make it converge to 0? $\endgroup$ Commented Oct 9, 2015 at 20:11
  • $\begingroup$ yep. $f(x) \approx \exp(-\frac2{3}x)$. This is a much stronger way to study asymptots. Could even be not horizontal, or not linear. $\endgroup$ Commented Oct 9, 2015 at 20:17
  • $\begingroup$ Right. Thanks for the help. $\endgroup$ Commented Oct 9, 2015 at 20:19

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .