Limit of a difference of cosines over x³
Evaluate the limit: $$ L = \lim_{x \to 0} \frac{\cos(xe^x)-\cos(xe^{-x})}{x^3}, $$ using the sum-to-product identity $$ \cos(xe^x)-\cos(xe^{-x}) = -2\sin\left(\frac{xe^x+xe^{-x}}{2}\right)\sin\left(\frac{xe^x-xe^{-x}}{2}\right). $$
Explanation of Method
Direct substitution gives $\frac{1-1}{0}=\frac00$. The given sum-to-product identity converts the difference of two cosines into a product of two sines, whose arguments simplify neatly: $\frac{xe^x+xe^{-x}}{2}=x\cosh x$ and $\frac{xe^x-xe^{-x}}{2}=x\sinh x$, using the definitions of hyperbolic cosine and sine. Since $\cosh x\to1$ and $\sinh x\sim x$ for small $x$, these arguments behave like $x$ and $x^2$ respectively, and the corresponding sines inherit that same leading-order behavior, producing a numerator of order $x^3$ that matches the denominator exactly.
- Rewrite the two arguments using $\cosh x=\frac{e^x+e^{-x}}2$ and $\sinh x=\frac{e^x-e^{-x}}2$.
- Approximate $x\cosh x$ and $x\sinh x$ for small $x$.
- Apply the small-angle approximation to each sine factor.
- Combine and divide by $x^3$ to find the limit.
Step 1 · Identify the arguments as $x\cosh x$ and $x\sinh x$
By definition, $\cosh x = \dfrac{e^x+e^{-x}}2$ and $\sinh x = \dfrac{e^x-e^{-x}}2$, so:
$$ \frac{xe^x+xe^{-x}}{2} = x\cdot\frac{e^x+e^{-x}}{2} = x\cosh x, \qquad \frac{xe^x-xe^{-x}}{2} = x\cdot\frac{e^x-e^{-x}}{2} = x\sinh x. $$
So the identity becomes:
$$ \cos(xe^x)-\cos(xe^{-x}) = -2\sin(x\cosh x)\sin(x\sinh x). $$
Step 2 · Approximate $x\cosh x$ and $x\sinh x$ for small $x$
Using the Taylor expansions $\cosh x = 1+\dfrac{x^2}2+O(x^4)$ and $\sinh x = x+\dfrac{x^3}6+O(x^5)$:
$$ x\cosh x = x + \frac{x^3}{2}+O(x^5) = x + O(x^3), \qquad x\sinh x = x^2+\frac{x^4}{6}+O(x^6) = x^2+O(x^4). $$
Step 3 · Apply the small-angle approximation to each sine
Since both arguments tend to $0$ as $x\to0$, use $\sin\theta\approx\theta$ for small $\theta$:
$$ \sin(x\cosh x) \approx x+O(x^3), \qquad \sin(x\sinh x) \approx x^2+O(x^4). $$
Step 4 · Combine and divide by $x^3$
Multiplying the two approximations:
$$ \sin(x\cosh x)\sin(x\sinh x) \approx \left(x+O(x^3)\right)\left(x^2+O(x^4)\right) = x^3+O(x^5). $$
So the numerator is:
$$ \cos(xe^x)-\cos(xe^{-x}) = -2\left(x^3+O(x^5)\right) = -2x^3+O(x^5). $$
Dividing by $x^3$ and taking the limit:
$$ L = \lim_{x\to0}\frac{-2x^3+O(x^5)}{x^3} = \lim_{x\to0}\left(-2+O(x^2)\right) = -2. $$
Summary Table
| Quantity | Approximation for small $x$ |
|---|---|
| $x\cosh x$ | $x+O(x^3)$ |
| $x\sinh x$ | $x^2+O(x^4)$ |
| $\sin(x\cosh x)\sin(x\sinh x)$ | $x^3+O(x^5)$ |
| $\cos(xe^x)-\cos(xe^{-x})$ | $-2x^3+O(x^5)$ |
Additional Notes
- Recognizing $\frac{e^x\pm e^{-x}}2$ as $\cosh x$ and $\sinh x$ is the key simplification here — without it, the arguments $\frac{xe^x+xe^{-x}}2$ and $\frac{xe^x-xe^{-x}}2$ look far more complicated than they really are.
- The two arguments vanish at different rates — $x\cosh x\sim x$ (linear) while $x\sinh x\sim x^2$ (quadratic) — and it is exactly this mismatch (one order $x$, one order $x^2$) that produces the needed order $x^3$ in the product, matching the denominator.
- This problem is a good illustration of combining several standard tools at once: a trigonometric identity to split the difference, hyperbolic function definitions to simplify the arguments, and small-angle approximations to finish the computation — no L'Hôpital's Rule was needed at any point.
Result
The value of the limit is: