Note that lim x → ∞ ( x + 1 ) ( x + 3 ) e − x = lim x → ∞ ( x + 1 ) ( x + 3 ) e x {\displaystyle \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 → ∞ . {\displaystyle x\to \infty .} Applying the L'Hopital's Rule stated in the hint twice, we have lim x → ∞ ( x + 1 ) ( x + 3 ) e − x = lim x → ∞ ( x + 1 ) ( x + 3 ) e x = lim x → ∞ x 2 + 4 x + 3 e x = lim x → ∞ 2 x + 4 e x = lim x → ∞ 2 e x = 0. {\displaystyle \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 0 {\displaystyle \color {blue}0} .
