Limit of (sinh x)² over ln(cosh 3x)
Evaluate the limit: $$ L = \lim_{x \to 0} \frac{\sinh^2x}{\ln(\cosh3x)}. $$
Explanation of Method
Direct substitution gives $\frac{0}{\ln1}=\frac00$. Both hyperbolic functions have well-known small-argument rates: $\sinh x\sim x$, so $\sinh^2x\sim x^2$; and $\cosh(3x)\sim1+\frac{(3x)^2}2$, so $\ln(\cosh3x)\sim\frac{(3x)^2}2=\frac{9x^2}2$ (using $\ln(1+v)\approx v$). Both numerator and denominator vanish at the same order, $x^2$, so their ratio tends to the ratio of their coefficients.
- Expand $\sinh x$ and square it to find the leading behavior of the numerator.
- Expand $\cosh(3x)$ and take its logarithm to find the leading behavior of the denominator.
- Divide the two leading behaviors and take the limit.
Step 1 · Expand the numerator
The Taylor series for $\sinh x$ near $0$ is $\sinh x = x+\dfrac{x^3}6+O(x^5)$, so:
$$ \sinh^2x = \left(x+\frac{x^3}6+O(x^5)\right)^2 = x^2+\frac{x^4}{3}+O(x^6). $$
Step 2 · Expand the denominator
The Taylor series for $\cosh x$ near $0$ is $\cosh x=1+\dfrac{x^2}2+O(x^4)$, so for $\cosh(3x)$:
$$ \cosh(3x) = 1+\frac{(3x)^2}{2}+O(x^4) = 1+\frac{9x^2}{2}+O(x^4). $$
Using $\ln(1+v)\approx v$ for small $v=\dfrac{9x^2}2+O(x^4)$:
$$ \ln(\cosh3x) \approx \frac{9x^2}{2}+O(x^4). $$
Step 3 · Divide and take the limit
Substituting both expansions:
$$ L = \lim_{x\to0} \frac{x^2+\frac{x^4}3+O(x^6)}{\frac{9x^2}2+O(x^4)}. $$
Dividing numerator and denominator by $x^2$:
$$ L = \lim_{x\to0} \frac{1+\frac{x^2}3+O(x^4)}{\frac92+O(x^2)} = \frac{1}{9/2} = \frac29. $$
Summary Table
| Quantity | Leading behavior for small $x$ |
|---|---|
| $\sinh^2x$ | $x^2$ |
| $\ln(\cosh3x)$ | $\dfrac{9x^2}2$ |
| Ratio | $\dfrac29$ |
Additional Notes
- The same limit can be derived using only the exponential definitions $\sinh x=\frac{e^x-e^{-x}}2$ and $\cosh x=\frac{e^x+e^{-x}}2$, expanding each exponential to the needed order — this gives an independent check of the same result, $\frac29$, without invoking the hyperbolic Taylor series by name.
- The coefficient $9$ in the denominator comes entirely from squaring the scaling factor $3$ inside $\cosh(3x)$ — replacing $3x$ with $cx$ for a general constant $c$ would give $\ln(\cosh(cx))\sim\frac{c^2x^2}2$, and hence $\lim_{x\to0}\frac{\sinh^2x}{\ln(\cosh(cx))}=\frac2{c^2}$.
- This is another instance of the general principle used throughout these problems: when both numerator and denominator vanish at the same order in $x$, only the coefficients of that matching leading order determine the limit — everything else can be discarded.
Result
The value of the limit is: