`lim_(xto0)(sinx^(4)-x^(4)cosx^(4))/(x^(4)(e^(2x^(4))-1-2x^(4)))` is

Options

A. 0
B. `-1/6`
C. `1/6`
D. Does not exist

Answer

To solve the limit \( \lim_{x \to 0} \frac{\sin(x^4) - x^4 \cos(x^4)}{x^4 (e^{2x^4} - 1 - 2x^4)} \), we will use Taylor series expansions for the functions involved.

### Step 1: Expand \(\sin(x^4)\) and \(\cos(x^4)\)
The Taylor series expansion for \(\sin(x)\) around \(x=0\) is:
\[ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots \]
Thus, substituting \(x^4\) for \(x\):
\[ \sin(x^4) = x^4 - \frac{(x^4)^3}{3!} + O(x^{16}) = x^4 - \frac{x^{12}}{6} + O(x^{16}) \]

The Taylor series expansion for \(\cos(x)\) around \(x=0\) is:
\[ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots \]
Thus, substituting \(x^4\) for \(x\):
\[ \cos(x^4) = 1 - \frac{(x^4)^2}{2} + O(x^{16}) = 1 - \frac{x^8}{2} + O(x^{16}) \]

### Step 2: Substitute expansions into the limit
Now we substitute these expansions into the limit:
\[ \sin(x^4) - x^4 \cos(x^4) = \left(x^4 - \frac{x^{12}}{6} + O(x^{16})\right) - x^4\left(1 - \frac{x^8}{2} + O(x^{16})\right) \]
This simplifies to:
\[ = x^4 - \frac{x^{12}}{6} - x^4 + \frac{x^{12}}{2} + O(x^{16}) = \left(\frac{1}{2} - \frac{1}{6}\right)x^{12} + O(x^{16}) = \frac{1}{3}x^{12} + O(x^{16}) \]

### Step 3: Expand \(e^{2x^4}\)
The Taylor series expansion for \(e^x\) is:
\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]
Thus, substituting \(2x^4\) for \(x\):
\[ e^{2x^4} = 1 + 2x^4 + \frac{(2x^4)^2}{2} + O(x^{12}) = 1 + 2x^4 + 2x^8 + O(x^{12}) \]
Now, substituting this into the denominator:
\[ e^{2x^4} - 1 - 2x^4 = (1 + 2x^4 + 2x^8 + O(x^{12})) - 1 - 2x^4 = 2x^8 + O(x^{12}) \]

### Step 4: Substitute into the limit expression
Now we substitute back into the limit:
\[ \lim_{x \to 0} \frac{\frac{1}{3}x^{12} + O(x^{16})}{x^4(2x^8 + O(x^{12}))} \]
This simplifies to:
\[ = \lim_{x \to 0} \frac{\frac{1}{3}x^{12}}{2x^{12} + O(x^{16})} = \lim_{x \to 0} \frac{\frac{1}{3}}{2 + O(x^4)} = \frac{1}{3 \cdot 2} = \frac{1}{6} \]

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