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

As $x\to +\infty$, the rate at which the exponential function $e^{-x}$ decays to zero is much faster than the rate at which the two linear functions $x+2$ and $x+1$ grow; therefore

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

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

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

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

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

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