# Science:Math Exam Resources/Courses/MATH100/December 2015/Question 10 (c)/Solution 1

When ${\displaystyle x\rightarrow \infty }$, The rate at which the exponential function ${\displaystyle e^{-x}}$ decays to zero is much faster than the rate at which the two linear functions ${\displaystyle x+2}$ and ${\displaystyle x+1}$ increase, therefore

${\displaystyle {\underset {x\rightarrow \pm \infty }{\lim(}}x+2)(x+1)e^{-x}=0}$

Indeed, more rigorously, it follows from L'Hospital's rule:

${\displaystyle {\underset {x\rightarrow \pm \infty }{\lim(}}x+2)(x+1)e^{-x}={\underset {x\rightarrow \pm \infty }{\lim }}{\frac {x^{2}+3x+2}{e^{x}}}={\underset {x\rightarrow \pm \infty }{\lim }}{\frac {2x+3}{e^{x}}}={\underset {x\rightarrow \pm \infty }{\lim }}{\frac {2}{e^{x}}}=0.}$

On the other hand, when ${\displaystyle x\rightarrow -\infty }$, using the graph of the functions we can easily see that

${\displaystyle {\underset {x\rightarrow -\infty }{\lim(}}x+2)(x+1)=+\infty }$ and ${\displaystyle {\underset {x\rightarrow -\infty }{\lim }}e^{-x}=+\infty }$.

Therefore, ${\displaystyle f(x)}$ has a horizontal asymtote only when ${\displaystyle x\rightarrow +\infty }$ which is at the very end of positive x-axis.

Horizontal Asymptote: ${\displaystyle \color {blue}y=0}$