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Squeeze Theorem

AP Calculus AB· difficulty 3/5

Evaluate lim⁡x→01−cos⁡xx\displaystyle\lim_{x \to 0}\dfrac{1 - \cos x}{x}.

  • A

    −1-1

  • B

    00

    check_circle
  • C

    12\tfrac{1}{2}

  • D

    11

Explanation

Standard limit: lim⁡x→0(1−cos⁡x)/x=0\lim_{x\to 0}(1-\cos x)/x = 0. (Multiply by conjugate: 1−cos⁡2xx(1+cos⁡x)=sin⁡2xx(1+cos⁡x)→0⋅1/2=0\dfrac{1-\cos^2 x}{x(1+\cos x)} = \dfrac{\sin^2 x}{x(1+\cos x)} \to 0 \cdot 1/2 = 0.)

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