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

As ${\displaystyle x\to +\infty }$, 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 ${\displaystyle x+1}$ grow; therefore

${\displaystyle \lim _{x\to +\infty }(x+2)(x+1)e^{-x}=0.}$

More rigorously, we can apply l'Hospital's rule twice to compute that

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

On the other hand, as ${\displaystyle x\to -\infty }$, we have ${\displaystyle \lim _{x\to -\infty }x+2=-\infty }$, ${\displaystyle \lim _{x\to -\infty }x+1=-\infty }$, and ${\displaystyle \lim _{x\to -\infty }e^{-x}=+\infty }$, so

${\displaystyle \lim _{x\to -\infty }(x+2)(x+1)e^{-x}=-\infty \cdot -\infty \cdot +\infty =+\infty .}$

Therefore the only asymptote of the function is the line ${\displaystyle {\color {blue}y=0}}$.