The value of `lim_(x to 0) (cos(sin x) - cos x)/x^(4)` is equal to:

Options

A. `1/5`
B. `1/6`
C. `1/4`
D. `1/2`

Answer

To find the limit

\[ L = \lim_{x \to 0} \frac{\cos(\sin x) - \cos x}{x^4}, \]

we can use Taylor series expansions for \(\cos(\sin x)\) and \(\cos x\) around \(x = 0\).

### Step 1: Taylor Series Expansion
The Taylor series expansion of \(\cos x\) around \(0\) is:

\[ \cos x = 1 - \frac{x^2}{2} + \frac{x^4}{24} - \frac{x^6}{720} + O(x^8). \]

For \(\sin x\), the Taylor series expansion is:

\[ \sin x = x - \frac{x^3}{6} + O(x^5). \]

### Step 2: Substitute \(\sin x\) into \(\cos(\sin x)\)
Now, substituting \(\sin x\) into the expansion of \(\cos\):

\[ \cos(\sin x) = \cos\left(x - \frac{x^3}{6} + O(x^5)\right). \]

Using the Taylor series for \(\cos\):

\[ \cos(\sin x) = 1 - \frac{\left(x - \frac{x^3}{6}\right)^2}{2} + O\left(\left(x - \frac{x^3}{6}\right)^4\right). \]

Calculating \(\left(x - \frac{x^3}{6}\right)^2\):

\[ \left(x - \frac{x^3}{6}\right)^2 = x^2 - \frac{x^4}{3} + O(x^6). \]

Thus,

\[ \cos(\sin x) = 1 - \frac{x^2 - \frac{x^4}{3}}{2} + O(x^4) = 1 - \frac{x^2}{2} + \frac{x^4}{6} + O(x^4). \]

### Step 3: Calculate \(\cos(\sin x) - \cos x\)
Now we compute:

\[ \cos(\sin x) - \cos x = \left(1 - \frac{x^2}{2} + \frac{x^4}{6}\right) - \left(1 - \frac{x^2}{2} + \frac{x^4}{24}\right). \]

This simplifies to:

\[ \cos(\sin x) - \cos x = \frac{x^4}{6} - \frac{x^4}{24} = \frac{4x^4 - x^4}{24} = \frac{3x^4}{24} = \frac{x^4}{8}. \]

### Step 4: Substitute back into the limit
Now substitute this back into the limit:

\[ L = \lim_{x \to 0} \frac{\frac{x^4}{8}}{x^4} = \lim_{x \to 0} \frac{1}{8} = \frac{1}{8}. \]

### Final Result
Thus, the value of the limit is:

\[ \boxed{\frac{1}{8}}. \]