The limit arises in the evaluation of certain improper integrals

MATHEMATICS · Single correct · AP · Calculus AB · Limits and Continuity · Limits and Continuity

Problem

The limit

\displaystyle \lim_{b\to\infty} b e^{-b}

arises in the evaluation of certain improper integrals. Which of the following best explains why this limit equals

0

? (a) Because

e^{-b}\to 0

b\to\infty

, multiplying it by any quantity gives

0

, so

b e^{-b}\to 0

. (b) Since the linear term

b

increases without bound, it eventually overrides the decay of

e^{-b}

, causing

b e^{-b}

to diverge to

\infty

. (c) An exponential function decays faster than any polynomial grows, so

e^{-b}

dominates

b

and their product approaches

0

. (d) Applying L’Hôpital’s Rule once gives

\displaystyle\lim_{b\to\infty}\tfrac{b}{e^{b}}=\tfrac{1}{e^{b}}\to 0

, so the original limit must equal

1