`lim_(xto0) (1)/(x^12){1-cos (x^2/2)-cos (x^4/4)+cos (x^2/2) cos (x^4/4)} ` is equal to

Options

A. `(1)/(32)`
B. `(1)/(256)`
C. ` (1)/(16)`
D. `-(1)/(256)`

Answer

To solve the limit problem

\[ \lim_{x \to 0} \frac{1}{x^{12}} \left( 1 - \cos\left(\frac{x^2}{2}\right) - \cos\left(\frac{x^4}{4}\right) + \cos\left(\frac{x^2}{2}\right) \cos\left(\frac{x^4}{4}\right) \right), \]

we will follow these steps:

### Step 1: Rewrite the expression
We start by rewriting the expression inside the limit:

\[ L = \lim_{x \to 0} \frac{1}{x^{12}} \left( 1 - \cos\left(\frac{x^2}{2}\right) - \cos\left(\frac{x^4}{4}\right) + \cos\left(\frac{x^2}{2}\right) \cos\left(\frac{x^4}{4}\right) \right). \]

### Step 2: Use trigonometric identities
Using the identity \(1 - \cos A = 2 \sin^2\left(\frac{A}{2}\right)\), we can express the cosines:

\[ 1 - \cos\left(\frac{x^2}{2}\right) = 2 \sin^2\left(\frac{x^2}{4}\right), \]
\[ 1 - \cos\left(\frac{x^4}{4}\right) = 2 \sin^2\left(\frac{x^4}{8}\right). \]

Substituting these into our limit gives:

\[ L = \lim_{x \to 0} \frac{1}{x^{12}} \left( 2 \sin^2\left(\frac{x^2}{4}\right) - 2 \sin^2\left(\frac{x^4}{8}\right) + \cos\left(\frac{x^2}{2}\right) \cos\left(\frac{x^4}{4}\right) \right). \]

### Step 3: Simplify the expression
Now we can factor out the common terms:

\[ L = 2 \lim_{x \to 0} \frac{1}{x^{12}} \left( \sin^2\left(\frac{x^2}{4}\right) - \sin^2\left(\frac{x^4}{8}\right) + \cos\left(\frac{x^2}{2}\right) \cos\left(\frac{x^4}{4}\right) \right). \]

### Step 4: Analyze the limit
As \(x \to 0\), both \(\sin\left(\frac{x^2}{4}\right)\) and \(\sin\left(\frac{x^4}{8}\right)\) approach 0. We can use the small angle approximation \(\sin z \approx z\) for small \(z\):

\[ \sin\left(\frac{x^2}{4}\right) \approx \frac{x^2}{4}, \quad \sin\left(\frac{x^4}{8}\right) \approx \frac{x^4}{8}. \]

Thus, we have:

\[ \sin^2\left(\frac{x^2}{4}\right) \approx \left(\frac{x^2}{4}\right)^2 = \frac{x^4}{16}, \quad \sin^2\left(\frac{x^4}{8}\right) \approx \left(\frac{x^4}{8}\right)^2 = \frac{x^8}{64}. \]

### Step 5: Substitute back into the limit
Substituting these approximations back into our limit gives:

\[ L = 2 \lim_{x \to 0} \frac{1}{x^{12}} \left( \frac{x^4}{16} - \frac{x^8}{64} + 1 \right). \]

### Step 6: Evaluate the limit
Now we can simplify:

\[ L = 2 \lim_{x \to 0} \left( \frac{x^4}{16x^{12}} - \frac{x^8}{64x^{12}} + \frac{1}{x^{12}} \right) = 2 \lim_{x \to 0} \left( \frac{1}{16x^8} - \frac{1}{64x^4} + \frac{1}{x^{12}} \right). \]

As \(x \to 0\), the dominant term is \(\frac{1}{x^{12}}\), leading to:

\[ L \to 2 \cdot \infty = \infty. \]

### Final Result
Thus, the limit diverges, and we conclude that:

\[ \lim_{x \to 0} \frac{1}{x^{12}} \left( 1 - \cos\left(\frac{x^2}{2}\right) - \cos\left(\frac{x^4}{4}\right) + \cos\left(\frac{x^2}{2}\right) \cos\left(\frac{x^4}{4}\right) \right) = 0. \]