Solve `lim_(x to0)(1-cos(1-cosx))/(sin^(4)x)`

Answer

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

### Step 1: Rewrite the Limit
We start by rewriting the limit expression:
\[ \lim_{x \to 0} \frac{1 - \cos(1 - \cos x)}{\sin^4 x} \]

### Step 2: Use the Identity for Cosine
Recall the limit identity:
\[ \lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2} \]
We will apply this identity to the term \(1 - \cos(1 - \cos x)\). To do this, we first need to analyze \(1 - \cos x\) as \(x\) approaches 0.

### Step 3: Expand \(1 - \cos x\)
Using the Taylor series expansion for \(\cos x\) around \(x = 0\):
\[ \cos x \approx 1 - \frac{x^2}{2} + O(x^4) \]
Thus,
\[ 1 - \cos x \approx \frac{x^2}{2} \]

### Step 4: Substitute in the Limit
Substituting \(1 - \cos x\) in our limit gives:
\[ 1 - \cos(1 - \cos x) \approx 1 - \cos\left(\frac{x^2}{2}\right) \]

### Step 5: Apply the Cosine Identity Again
Now we apply the cosine limit identity again:
\[ 1 - \cos\left(\frac{x^2}{2}\right) \approx \frac{1}{2}\left(\frac{x^2}{2}\right)^2 = \frac{x^4}{8} \]

### Step 6: Substitute Back into the Limit
Now we substitute this back into our limit:
\[ \lim_{x \to 0} \frac{\frac{x^4}{8}}{\sin^4 x} \]

### Step 7: Use the Sine Limit Identity
Using the identity:
\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \implies \sin x \approx x \text{ as } x \to 0 \]
Thus,
\[ \sin^4 x \approx x^4 \]

### Step 8: Final Limit Calculation
Now substituting this into our limit:
\[ \lim_{x \to 0} \frac{\frac{x^4}{8}}{x^4} = \lim_{x \to 0} \frac{1}{8} = \frac{1}{8} \]

### Final Answer
Thus, the limit evaluates to:
\[ \frac{1}{8} \]