# Find the horizontal asymptotes

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.

• Antonio, why did you edit it like that? Now it doesn't make any sense. – Daniel Waleniak Oct 9 '15 at 19:42
• 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 $x\to+\infty$. – John Molokach Oct 9 '15 at 19:46
• My professor doesn't let us use L'Hopital's rule unfortunately. Thanks for the hint. – Daniel Waleniak Oct 9 '15 at 19:49
• 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. – André Nicolas Oct 9 '15 at 19:50
• @DanielWaleniak: a positive number (greater than $1$) with an exponent tending toward negative infinity will go to $0$. – Clayton Oct 9 '15 at 20:02

Hints: what is the first order approximation of $3+4x^2$ towards infinity ? what about $2-6x$ ?
• I got $\frac {4}{0}$ in the exponent when I substituted x for infinity. – Daniel Waleniak Oct 9 '15 at 19:58
• Would it be something like $\frac {0+4}{0-6x}$ in the exponent giving $\frac{-2x}{3}$ which would make it converge to 0? – Daniel Waleniak Oct 9 '15 at 20:11
• 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. – Fabrice NEYRET Oct 9 '15 at 20:17