Consider . Which statement best explains why does not…

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

Problem

Consider

f(x)=\sin\!\left(\dfrac{1}{x}\right)

. Which statement best explains why

\displaystyle \lim_{x\to 0}f(x)

does not exist? (a) The function has a vertical asymptote at

x=0

, so its values grow without bound as

x

approaches

0

. (b) Because

\displaystyle\lim_{x\to0}\tfrac1x

does not exist, the continuity of the sine function cannot produce a limit for

f(x)

. (c) The function oscillates infinitely often between

-1

and

1

x

approaches

0

, preventing it from settling on a single value. (d) A limit cannot exist when the function is not defined at the point in question, and

f(x)

is undefined at

x=0