AP Calculus AB and BC

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

The limit of the natural log of cosine x, over x squared, as x approaches 0 is negative one half. The form is 0 over 0, so L'Hopital's rule applies: the first pass gives negative tangent x over 2x, still 0 over 0, and the second gives negative secant squared x over 2.

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

Settled by L'Hopital's rule applied twice.

Two passes of L'Hopital

At x=0x = 0 the numerator is ln⁡1=0\ln 1 = 0 and the denominator is 00. Differentiating each, and remembering that ddxln⁡(cos⁡x)=−sin⁡xcos⁡x=−tan⁡x\frac{d}{dx}\ln(\cos x) = \frac{-\sin x}{\cos x} = -\tan x:

lim⁡x→0ln⁡(cos⁡x)x2=lim⁡x→0−tan⁡x2x\lim_{x \to 0}\frac{\ln\left(\cos x\right)}{x^{2}} = \lim_{x \to 0}\frac{-\tan x}{2x}

That is 00\frac{0}{0} again, so run the rule a second time. The derivative of −tan⁡x-\tan x is −sec⁡2x-\sec^{2}x and the derivative of 2x2x is 22.

lim⁡x→0−tan⁡x2x=lim⁡x→0−sec⁡2x2=−12\lim_{x \to 0}\frac{-\tan x}{2x} = \lim_{x \to 0}\frac{-\sec^{2}x}{2} = -\frac{1}{2}

The minus sign is real

For xx near 00 but not equal to it, 0<cos⁡x<10 < \cos x < 1, so ln⁡(cos⁡x)\ln(\cos x) is negative. The denominator x2x^{2} is positive on both sides. A negative over a positive is negative, which is the sign check to run before you trust the algebra.

A second confirmation comes from the standard limit cos⁡x−1x2→−12\frac{\cos x - 1}{x^{2}} \to -\frac{1}{2} together with ln⁡(1+u)≈u\ln(1+u) \approx u for small uu. Put u=cos⁡x−1u = \cos x - 1 and the same −12-\frac{1}{2} appears without any differentiation.

Sizing it up

Near the origin ln(cos x) is roughly minus one half x squared. So the graph of ln(cos x) sits just under the axis and opens downward, matching a limit of -1/2 for the quotient.

The mistakes students make

The chain rule inside the logarithm is where nearly all the lost marks sit.

  • Answering 12\frac{1}{2} by dropping the minus that ddxln⁡(cos⁡x)=−tan⁡x\frac{d}{dx}\ln(\cos x) = -\tan x carries.
  • Differentiating ln⁡(cos⁡x)\ln(\cos x) as 1cos⁡x\frac{1}{\cos x} and losing the inner −sin⁡x-\sin x. That leaves sec⁡x2x\frac{\sec x}{2x}, which makes the limit look infinite.
  • Stopping after one pass at −tan⁡x2x\frac{-\tan x}{2x} and calling it 00. It is still 00\frac{0}{0} and needs the rule again.

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 ln(cos x)/x^2 as x approaches 0?

It is −12-\frac{1}{2}.

Why is the answer negative?

Because cos⁡x<1\cos x < 1 for xx near 00, so ln⁡(cos⁡x)<0\ln(\cos x) < 0, while x2>0x^{2} > 0 on both sides of the origin.

Can I do this without L'Hopital's rule?

Yes. Write ln⁡(cos⁡x)=ln⁡(1+(cos⁡x−1))≈cos⁡x−1\ln(\cos x) = \ln\left(1 + (\cos x - 1)\right) \approx \cos x - 1, then use lim⁡x→0cos⁡x−1x2=−12\lim_{x \to 0}\frac{\cos x - 1}{x^{2}} = -\frac{1}{2}.