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Squeeze Theorem

AP Calculus AB· difficulty 4/5

lim⁡x→0x2sin⁡(1/x)=\displaystyle \lim_{x\to 0} x^2 \sin(1/x) =

  • A

    00

    check_circle
  • B

    Does not exist

  • C

    ∞\infty

  • D

    11

Explanation

Squeeze: −x2≤x2sin⁡(1/x)≤x2-x^2 \le x^2\sin(1/x) \le x^2. Both bounds →0\to 0 as x→0x\to 0.

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