Limit of ln(cos ax) over ln(cos bx)
Evaluate the limit, for constants $a$ and $b\neq0$: $$ L = \lim_{x \to 0} \frac{\ln\cos(ax)}{\ln\cos(bx)}. $$
Explanation of Method
Both $\ln\cos(ax)$ and $\ln\cos(bx)$ tend to $\ln(1)=0$ as $x\to0$, giving $\frac00$. Since $1-\cos(kx)\sim\frac{(kx)^2}2$ for small $x$ (a standard small-angle rate seen elsewhere on this site), and $\ln(1-y)\approx-y$ for small $y$, each logarithm inherits the same quadratic rate, differing only by the square of its respective coefficient, $a^2$ or $b^2$. Dividing the two eliminates the common factor of $x^2$ entirely, leaving a constant ratio.
- Approximate $\cos(ax)$ and $\cos(bx)$ using the Taylor expansion of cosine.
- Approximate $\ln\cos(ax)$ and $\ln\cos(bx)$ using $\ln(1-y)\approx-y$.
- Divide the two approximations and simplify.
Step 1 · Approximate the cosines
Using $\cos(kx)\approx1-\dfrac{(kx)^2}2$ for small $x$:
$$ \cos(ax) \approx 1-\frac{(ax)^2}{2}, \qquad \cos(bx) \approx 1-\frac{(bx)^2}{2}. $$
Step 2 · Approximate the logarithms
Using $\ln(1-y)\approx-y$ for small $y$:
$$ \ln\cos(ax) \approx -\frac{(ax)^2}{2}, \qquad \ln\cos(bx) \approx -\frac{(bx)^2}{2}. $$
Step 3 · Divide and simplify
Substituting these approximations into the limit:
$$ L = \lim_{x\to0} \frac{-\frac{(ax)^2}{2}}{-\frac{(bx)^2}{2}} = \lim_{x\to0}\frac{(ax)^2}{(bx)^2} = \lim_{x\to0}\frac{a^2x^2}{b^2x^2} = \frac{a^2}{b^2}. $$
Since the ratio $\dfrac{a^2}{b^2}$ no longer depends on $x$ at all (the factor of $x^2$ cancelled completely), this is already the value of the limit.
Summary Table
| Quantity | Approximation for small $x$ |
|---|---|
| $\ln\cos(ax)$ | $-\dfrac{a^2x^2}2$ |
| $\ln\cos(bx)$ | $-\dfrac{b^2x^2}2$ |
| Ratio | $\dfrac{a^2}{b^2}$ |
Additional Notes
- Since both numerator and denominator vanish at exactly the same order ($x^2$) with coefficients that only differ through the constants $a$ and $b$, the ratio simplifies to a pure constant $\frac{a^2}{b^2}$ — no residual dependence on $x$ survives, which is a hallmark of comparing two quantities with matching leading-order behavior.
- The result is valid even if $a=0$: then $\cos(ax)=\cos(0)=1$ identically, so $\ln\cos(ax)=0$ for all $x$, and the ratio is identically $0$, matching the formula $\frac{a^2}{b^2}=\frac0{b^2}=0$.
- This limit could equally be evaluated with L'Hôpital's Rule (differentiating $\ln\cos(ax)$ and $\ln\cos(bx)$, which introduces factors of $\tan$, and then taking a further limit), but recognizing the shared quadratic rate via Taylor expansion is considerably faster.
Result
The value of the limit is: