AP Calculus AB and BC

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

The limit of cos x minus 1, over x squared, as x approaches 0 is negative one half. It is the negative of the standard limit for 1 minus cos x over x squared. Two passes of L'Hopital's rule get there, and the Maclaurin series explains why the answer is a half.

lim⁡x→0cos⁡x−1x2=−12\lim_{x \to 0} \frac{\cos x - 1}{x^{2}} = -\frac{1}{2}

Settled by L'Hopital's rule twice, or the Maclaurin series.

Two passes of L'Hopital

Substitution gives 00\frac{0}{0}, and one pass leaves −sin⁡x2x\frac{-\sin x}{2x}, which is 00\frac{0}{0} again.

lim⁡x→0cos⁡x−1x2  =H  lim⁡x→0−sin⁡x2x  =H  lim⁡x→0−cos⁡x2=−12\lim_{x \to 0}\frac{\cos x - 1}{x^{2}} \;\overset{\text{H}}{=}\; \lim_{x \to 0}\frac{-\sin x}{2x} \;\overset{\text{H}}{=}\; \lim_{x \to 0}\frac{-\cos x}{2} = -\frac{1}{2}

The series says it in one line

cos x = 1 - x^2/2 + x^4/24 - ..., so cos x - 1 behaves like -x^2/2 near the origin. Dividing by x^2 leaves -1/2, and the series also shows the next correction is of order x^2.

Order matters, not just the value

cos⁡x−1\cos x - 1 vanishes to SECOND order while sin⁡x\sin x vanishes to first. That is why cos⁡x−1x\frac{\cos x - 1}{x} tends to 00 while cos⁡x−1x2\frac{\cos x - 1}{x^{2}} tends to a nonzero constant, and why cos⁡x−1x3\frac{\cos x - 1}{x^{3}} has no limit at all.

The mistakes students make

  • Getting the sign backwards. cos⁡x−1\cos x - 1 is NEGATIVE near 00, so the limit is negative.
  • Stopping after one pass at −sin⁡x2x\frac{-\sin x}{2x} and reporting 00.
  • Confusing it with 1−cos⁡xx2=12\frac{1-\cos x}{x^{2}} = \frac{1}{2}, the same limit with the sign flipped.

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 (cos x - 1)/x^2 as x approaches 0?

It is −12-\frac{1}{2}.

Why two passes of L'Hopital?

Because the first pass leaves −sin⁡x2x\frac{-\sin x}{2x}, which is still 00\frac{0}{0}.

What if the denominator were x?

Then the limit is 00, because the numerator vanishes to second order and the denominator only to first.