Science:Math Exam Resources/Courses/MATH110/April 2016/Question 04 (c)/Solution 1

Note that
$\lim _{x\to \infty }(x+1)(x+3)e^{-x}=\lim _{x\to \infty }{\frac {(x+1)(x+3)}{e^{x}}}$ with both the top and the bottom approach to infinity as $x\to \infty .$
Applying the L'Hopital's Rule stated in the hint twice, we have
$\lim _{x\to \infty }(x+1)(x+3)e^{-x}=\lim _{x\to \infty }{\frac {(x+1)(x+3)}{e^{x}}}=\lim _{x\to \infty }{\frac {x^{2}+4x+3}{e^{x}}}=\lim _{x\to \infty }{\frac {2x+4}{e^{x}}}=\lim _{x\to \infty }{\frac {2}{e^{x}}}=0.$

Thus, the limit is $\color {blue}0$.