The value of `lim_(xrarr0)(16-16 cos (1-cos x))/(x^(4))` is _____________

Answer

To solve the limit \( \lim_{x \to 0} \frac{16 - 16 \cos(1 - \cos x)}{x^4} \), we can follow these steps:

### Step 1: Simplify the Expression
We start by factoring out the common term from the numerator:
\[ \lim_{x \to 0} \frac{16(1 - \cos(1 - \cos x))}{x^4} \]

### Step 2: Identify the Form of the Limit
As \( x \to 0 \), both \( 1 - \cos(1 - \cos x) \) and \( x^4 \) approach 0. Therefore, we have a \( \frac{0}{0} \) indeterminate form.

### Step 3: Use the Identity for \( 1 - \cos \)
We know that \( 1 - \cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right) \). We can apply this identity:
\[ 1 - \cos(1 - \cos x) = 2 \sin^2\left(\frac{1 - \cos x}{2}\right) \]
Thus, we rewrite the limit:
\[ \lim_{x \to 0} \frac{16 \cdot 2 \sin^2\left(\frac{1 - \cos x}{2}\right)}{x^4} \]

### Step 4: Expand \( 1 - \cos x \)
Using the Taylor series expansion for \( \cos x \) around \( x = 0 \):
\[ \cos x \approx 1 - \frac{x^2}{2} + O(x^4) \]
This gives us:
\[ 1 - \cos x \approx \frac{x^2}{2} \]
So,
\[ \frac{1 - \cos x}{2} \approx \frac{x^2}{4} \]

### Step 5: Substitute Back into the Limit
Now substituting this back into our limit:
\[ \sin^2\left(\frac{1 - \cos x}{2}\right) \approx \sin^2\left(\frac{x^2}{4}\right) \]
Using the small angle approximation \( \sin t \approx t \) when \( t \to 0 \):
\[ \sin\left(\frac{x^2}{4}\right) \approx \frac{x^2}{4} \]
Thus,
\[ \sin^2\left(\frac{x^2}{4}\right) \approx \left(\frac{x^2}{4}\right)^2 = \frac{x^4}{16} \]

### Step 6: Substitute into the Limit
Now substituting this back into our limit expression:
\[ \lim_{x \to 0} \frac{16 \cdot 2 \cdot \frac{x^4}{16}}{x^4} = \lim_{x \to 0} \frac{32 \cdot x^4}{16 \cdot x^4} = \lim_{x \to 0} 2 = 2 \]

### Final Answer
Thus, the value of the limit is:
\[ \boxed{2} \]