Second difference of cos(a+x) over x²
Evaluate the limit: $$ L = \lim_{x \to 0} \frac{\cos(a+2x) - 2\cos(a+x) + \cos a}{x^2}. $$
Explanation of Method
Direct substitution gives $\frac{\cos a - 2\cos a+\cos a}{0}=\frac00$. Expanding $\cos(a+2x)$ and $\cos(a+x)$ with the angle-addition formula and the Taylor series in $x$ (treating $a$ as a fixed constant) reduces the numerator to a single surviving term of order $x^2$, after all the constant and linear-in-$x$ pieces cancel. This particular combination, $f(a+2x)-2f(a+x)+f(a)$, is in fact a standard finite-difference approximation to $x^2f''(a)$ for any twice-differentiable $f$, which offers a second, calculus-based way to see the same result.
- Expand $\cos(a+2x)$ and $\cos(a+x)$ using the angle-addition formula and Taylor series in $x$.
- Combine the expansions and simplify the numerator.
- Divide by $x^2$ and take the limit.
- Cross-check using the general finite-difference identity for $f''(a)$.
Step 1 · Expand each cosine term
Using $\cos(a+h)=\cos a\cos h-\sin a\sin h$ together with $\cos h\approx1-\frac{h^2}2$ and $\sin h\approx h$ for small $h$:
$$ \cos(a+2x) = \cos a\cos(2x)-\sin a\sin(2x) \approx \cos a\left(1-2x^2\right)-\sin a(2x) = \cos a-2x\sin a-2x^2\cos a+O(x^3), $$
$$ \cos(a+x) = \cos a\cos x-\sin a\sin x \approx \cos a\left(1-\frac{x^2}2\right)-\sin a(x) = \cos a-x\sin a-\frac{x^2}2\cos a+O(x^3). $$
Step 2 · Combine and simplify the numerator
Substituting into the numerator:
$$ \cos(a+2x)-2\cos(a+x)+\cos a $$
$$ \approx \left[\cos a-2x\sin a-2x^2\cos a\right]-2\left[\cos a-x\sin a-\frac{x^2}2\cos a\right]+\cos a + O(x^3). $$
Grouping like terms: the constant terms give $\cos a-2\cos a+\cos a=0$; the linear-in-$x$ terms give $-2x\sin a-(-2x\sin a)=0$; the quadratic terms give $-2x^2\cos a-(-x^2\cos a)=-x^2\cos a$. So:
$$ \cos(a+2x)-2\cos(a+x)+\cos a \approx -x^2\cos a + O(x^3). $$
Step 3 · Divide by $x^2$ and take the limit
$$ L = \lim_{x\to0}\frac{-x^2\cos a+O(x^3)}{x^2} = \lim_{x\to0}\left(-\cos a+O(x)\right) = -\cos a. $$
Step 4 · Cross-check via the second-derivative identity
For any function $f$ with a continuous second derivative near $a$, the Taylor expansions $f(a+h)=f(a)+hf'(a)+\frac{h^2}2f''(a)+O(h^3)$ (with $h=x$ and $h=2x$) give, after the same cancellation of constant and linear terms:
$$ f(a+2x)-2f(a+x)+f(a) = x^2f''(a)+O(x^3), \qquad\text{so}\qquad \lim_{x\to0}\frac{f(a+2x)-2f(a+x)+f(a)}{x^2}=f''(a). $$
With $f(t)=\cos t$, $f''(t)=-\cos t$, so this identity directly gives $L=f''(a)=-\cos a$, confirming Step 3 without redoing the trigonometric expansion.
Summary Table
| Term order | Contribution to the numerator |
|---|---|
| Constant ($x^0$) | $\cos a-2\cos a+\cos a=0$ |
| Linear ($x^1$) | $-2x\sin a+2x\sin a=0$ |
| Quadratic ($x^2$) | $-2x^2\cos a+x^2\cos a=-x^2\cos a$ |
Additional Notes
- The general identity $\lim_{x\to0}\dfrac{f(a+2x)-2f(a+x)+f(a)}{x^2}=f''(a)$ holds for any twice-differentiable $f$ — it is a standard finite-difference formula used, for instance, in numerical approximations of second derivatives.
- This gives a fast way to answer similar problems: for $f(t)=\sin t$, the same identity immediately gives $\lim_{x\to0}\frac{\sin(a+2x)-2\sin(a+x)+\sin a}{x^2}=f''(a)=-\sin a$, without repeating the trigonometric expansion from scratch.
- The exact cancellation of both the constant and linear terms in Step 2 is not a coincidence specific to cosine — it is exactly what makes this combination of function values a valid second-derivative approximation for any smooth function, regardless of which function is used.
Result
The value of the limit is: