The value of `lim _(xto0) (cos (sin x )- cos x)/(x ^(4))` is equal to :

Options

A. `1/5`
B. `1/6`
C. `1/4`
D. `(1)/(12)`

Answer

To solve the limit \( \lim_{x \to 0} \frac{\cos(\sin x) - \cos x}{x^4} \), we can follow these steps:

### Step 1: Apply the Cosine Difference Formula
We start by using the trigonometric identity for the difference of cosines:
\[ \cos a - \cos b = -2 \sin\left(\frac{a + b}{2}\right) \sin\left(\frac{a - b}{2}\right) \]
In our case, let \( a = \sin x \) and \( b = x \). Thus, we have:
\[ \cos(\sin x) - \cos x = -2 \sin\left(\frac{\sin x + x}{2}\right) \sin\left(\frac{\sin x - x}{2}\right) \]
Substituting this into our limit gives:
\[ \lim_{x \to 0} \frac{-2 \sin\left(\frac{\sin x + x}{2}\right) \sin\left(\frac{\sin x - x}{2}\right)}{x^4} \]

### Step 2: Simplify the Limit
Next, we need to analyze \( \sin\left(\frac{\sin x - x}{2}\right) \). Using the Taylor series expansion for small \( x \):
\[ \sin x \approx x - \frac{x^3}{6} + O(x^5) \]
Thus,
\[ \sin x - x \approx -\frac{x^3}{6} + O(x^5) \]
This implies:
\[ \frac{\sin x - x}{2} \approx -\frac{x^3}{12} + O(x^5) \]
So,
\[ \sin\left(\frac{\sin x - x}{2}\right) \approx \frac{-x^3}{12} \]

### Step 3: Substitute Back into the Limit
Now substituting back into our limit:
\[ \lim_{x \to 0} \frac{-2 \sin\left(\frac{\sin x + x}{2}\right) \left(-\frac{x^3}{12}\right)}{x^4} \]
As \( x \to 0 \), \( \sin\left(\frac{\sin x + x}{2}\right) \to \sin(0) = 0 \). We can approximate it:
\[ \sin\left(\frac{\sin x + x}{2}\right) \approx \frac{\sin x + x}{2} \approx \frac{x + x}{2} = x \]
So we have:
\[ \lim_{x \to 0} \frac{-2 \cdot x \cdot \left(-\frac{x^3}{12}\right)}{x^4} = \lim_{x \to 0} \frac{2x^4/12}{x^4} = \lim_{x \to 0} \frac{2}{12} = \frac{1}{6} \]

### Final Answer
Thus, the value of the limit is:
\[ \boxed{\frac{1}{6}} \]