To evaluate the limit [tex]\(\lim_{x \to 0} \frac{x \cos(x) - \sin(x)}{4 x^3}\)[/tex], we can use series expansion (Maclaurin series) of the trigonometric functions [tex]\( \cos(x) \)[/tex] and [tex]\( \sin(x) \)[/tex].
First, let's recall the Maclaurin series expansions for [tex]\( \cos(x) \)[/tex] and [tex]\( \sin(x) \)[/tex] around [tex]\( x = 0 \)[/tex]:
[tex]\[
\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + O(x^6)
\][/tex]
[tex]\[
\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} + O(x^7)
\][/tex]
Here, [tex]\(O(x^n)\)[/tex] represents the higher-order terms which become negligible as [tex]\(x\)[/tex] approaches 0.
Now let's consider the numerator [tex]\(x \cos(x) - \sin(x)\)[/tex]:
[tex]\[
x \cos(x) = x \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} + O(x^6)\right) = x - \frac{x^3}{2} + \frac{x^5}{24} + O(x^7)
\][/tex]
[tex]\[
\sin(x) = x - \frac{x^3}{6} + \frac{x^5}{120} + O(x^7)
\][/tex]
Subtract these two series:
[tex]\[
x \cos(x) - \sin(x) = \left( x - \frac{x^3}{2} + \frac{x^5}{24} + O(x^7) \right) - \left( x - \frac{x^3}{6} + \frac{x^5}{120} + O(x^7) \right)
\][/tex]
[tex]\[
= x - \frac{x^3}{2} + \frac{x^5}{24} - x + \frac{x^3}{6} - \frac{x^5}{120} + O(x^7)
\][/tex]
[tex]\[
= \left( -\frac{x^3}{2} + \frac{x^3}{6} \right) + \left( \frac{x^5}{24} - \frac{x^5}{120} \right) + O(x^7)
\][/tex]
[tex]\[
= -\frac{3x^3}{6} + \frac{x^3}{6} + O(x^5)
\][/tex]
[tex]\[
= -\frac{2x^3}{6} + O(x^5)
\][/tex]
[tex]\[
= -\frac{x^3}{3} + O(x^5)
\][/tex]
Thus, for small values of [tex]\( x \)[/tex],
[tex]\[
x \cos(x) - \sin(x) \approx -\frac{x^3}{3}
\][/tex]
Now we substitute this into the limit:
[tex]\[
\lim_{x \to 0} \frac{x \cos(x) - \sin(x)}{4 x^3}
\][/tex]
[tex]\[
= \lim_{x \to 0} \frac{-\frac{x^3}{3}}{4 x^3}
\][/tex]
[tex]\[
= \lim_{x \to 0} \frac{-x^3}{12 x^3}
\][/tex]
[tex]\[
= \lim_{x \to 0} \frac{-1}{12}
\][/tex]
Since [tex]\(x^3\)[/tex] terms cancel out, the limit simply becomes
[tex]\[
= -\frac{1}{12}
\][/tex]
So, the correct answer is
(D) [tex]\(-\frac{1}{12}\)[/tex]
