To find the limit [tex]\(\lim_{x \rightarrow 0} \frac{\cos x - \cos 5x}{x^2}\)[/tex], let's proceed with a step-by-step solution.
1. Expression given:
[tex]\[
\lim_{x \rightarrow 0} \frac{\cos x - \cos 5x}{x^2}
\][/tex]
2. Using the trigonometric identity for the difference of cosines:
[tex]\[
\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right) \sin\left(\frac{A-B}{2}\right)
\][/tex]
In our case, [tex]\(A = x\)[/tex] and [tex]\(B = 5x\)[/tex]. Using the identity:
[tex]\[
\cos x - \cos 5x = -2 \sin\left(\frac{x + 5x}{2}\right) \sin\left(\frac{x - 5x}{2}\right) = -2 \sin\left(3x\right) \sin\left(-2x\right)
\][/tex]
Simplifying further, since [tex]\(\sin(-2x) = -\sin(2x)\)[/tex]:
[tex]\[
\cos x - \cos 5x = -2 \sin(3x)(-\sin(2x)) = 2 \sin(3x) \sin(2x)
\][/tex]
3. Substitute this back into the original limit:
[tex]\[
\lim_{x \rightarrow 0} \frac{2 \sin(3x) \sin(2x)}{x^2}
\][/tex]
4. Separate the constant factor [tex]\(2\)[/tex]:
[tex]\[
2 \lim_{x \rightarrow 0} \frac{\sin(3x) \sin(2x)}{x^2}
\][/tex]
5. Use the limits of trigonometric functions as [tex]\(x \to 0\)[/tex]:
- Recall the standard limit result [tex]\(\lim_{u \to 0} \frac{\sin(u)}{u} = 1\)[/tex].
We can rewrite the sine terms using [tex]\(u = 3x\)[/tex] and [tex]\(v = 2x\)[/tex]:
[tex]\[
\frac{\sin(3x)}{3x} \cdot 3 \quad \text{and} \quad \frac{\sin(2x)}{2x} \cdot 2
\][/tex]
6. Combine changes of variables:
[tex]\[
2 \lim_{x \rightarrow 0} \sin(3x) \cdot \frac{3}{3x} \cdot \sin(2x) \cdot \frac{2}{2x} \cdot x^2
\][/tex]
Simplifying:
[tex]\[
2 \lim_{x \rightarrow 0} \left(\frac{\sin(3x)}{3x} \cdot \frac{\sin(2x)}{2x} \cdot 6\right)
\][/tex]
7. Apply the limit results:
[tex]\[
\lim_{x \rightarrow 0} \frac{\sin(3x)}{3x} = 1 \quad \text{and} \quad \lim_{x \rightarrow 0} \frac{\sin(2x)}{2x} = 1
\][/tex]
8. Calculate the overall limit:
[tex]\[
2 \times 1 \times 1 \times 6 = 12
\][/tex]
So, the limit is:
[tex]\[
\boxed{12}
\][/tex]
