Squeeze Theorem

AP Calculus AB· difficulty 3/5

Evaluate limx0xsin(1/x)\displaystyle\lim_{x\to 0} x \sin(1/x).

  • A

    \infty

  • B

    00

    check_circle
  • C

    11

  • D

    Does not exist

Explanation

xsin(1/x)x|x \sin(1/x)| \le |x| since sin()1|\sin(\cdot)| \le 1. Squeeze theorem: limit is 00.

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