AP Calculus AB and BC

Limit of (1-cos x)/(x sin x) at 0 Is 1/2

The limit of 1 minus cos x, over x sin x, as x approaches 0 is one half. Divide top and bottom by x squared: the numerator becomes the standard limit worth one half and the denominator becomes sin x over x, worth 1.

lim⁡x→01−cos⁡xxsin⁡x=12\lim_{x \to 0} \frac{1-\cos x}{x\sin x} = \frac{1}{2}

Settled by dividing through by x squared.

Divide by x squared

1−cos⁡xxsin⁡x=1−cos⁡xx2sin⁡xx⟶1/21=12\frac{1-\cos x}{x\sin x} = \frac{\frac{1-\cos x}{x^{2}}}{\frac{\sin x}{x}} \longrightarrow \frac{1/2}{1} = \frac{1}{2}

Both pieces are limits you already know, so no differentiation is needed. L'Hopital works but takes two passes and messier algebra.

Orders explain the answer

1−cos⁡x1-\cos x vanishes to second order and xsin⁡xx\sin x does too, so the ratio is finite and nonzero. Matching orders is the quickest way to predict whether a 00\frac{0}{0} form will give 00, a constant, or infinity.

The mistakes students make

  • Cancelling 1−cos⁡x1-\cos x against sin⁡x\sin x using an identity incorrectly. They are not proportional near 00.
  • Answering 11 by using only the sine limit and ignoring the numerator's order.

Not sure which technique a limit wants?

The Limit Method Chooser walks the decision from direct substitution through factoring, the conjugate, and L'Hopital, and says why each one applies or fails.

Frequently asked questions

What is the limit of (1-cos x)/(x sin x) at 0?

It is 12\frac{1}{2}.

Do I need L'Hopital?

No. Dividing top and bottom by x2x^{2} reduces it to two standard limits.