The limit equals

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

Problem

The limit

\displaystyle \lim_{x \to 0} x^{2}\sin\!\left(\tfrac{1}{x}\right)

equals

0

. Which of the following statements best justifies this result? (a)

\sin\!\left(\tfrac{1}{x}\right)

oscillates infinitely fast near

x=0

, so its average value becomes

0

, forcing the product to be

0

. (b)

\sin\!\left(\tfrac{1}{x}\right)

is always between

-1

and

1

, so multiplying it by

x^{2}

, whose magnitude approaches

0

, forces the product toward

0

. (Squeeze Theorem) (c) Substituting

u=\tfrac{1}{x}

gives

u\to\infty

, so the limit equals

\displaystyle\lim_{u\to\infty}\tfrac{\sin u}{u^{2}}

, which is declared

0

by L’Hôpital’s Rule. (d) Because

x^{2}\to 0

much “faster” than

\sin\!\left(\tfrac{1}{x}\right)

can vary,

x^{2}

dominates the product, so the limit must be

0