AP Calculus ABhardmcq1 pt

limx0x2sin(1x)=\lim_{x\to 0} x^2\sin\left(\frac{1}{x}\right)=

A.11
B.Does not exist
C.00
D.\infty

Explanation

Core Concept

Since 1sin(1/x)1-1\le \sin(1/x)\le 1, multiplying by x20x^2\ge 0 gives x2x2sin(1/x)x2-x^2\le x^2\sin(1/x)\le x^2. Both bounds approach 00, so by the Squeeze Theorem the limit is 00.

Correct Answer

C00

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