`lim_(xto0)(sinx^(4)-x^(4)cosx^(4)+x^(20))/(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

\[ \lim_{x \to 0} \frac{\sin(x^4) - x^4 \cos(x^4) + x^{20}}{x^4 (e^{2x^4} - 1 - 2x^4)}, \]

we will use Taylor series expansions for the functions involved.

### Step 1: Expand \(\sin(x^4)\) and \(\cos(x^4)\)

Using Taylor series, we have:
- \(\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots\)
- \(\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots\)

Thus, for \(\sin(x^4)\):
\[ \sin(x^4) = x^4 - \frac{x^{12}}{6} + O(x^{20}), \]
and for \(\cos(x^4)\):
\[ \cos(x^4) = 1 - \frac{x^8}{2} + O(x^{16}). \]

### Step 2: Substitute the expansions into the limit

Now substituting these expansions into the limit expression:
\[ \sin(x^4) - x^4 \cos(x^4) + x^{20} = \left(x^4 - \frac{x^{12}}{6} + O(x^{20})\right) - x^4 \left(1 - \frac{x^8}{2} + O(x^{16})\right) + x^{20}. \]

This simplifies to:
\[ x^4 - \frac{x^{12}}{6} - x^4 + \frac{x^{12}}{2} + x^{20} + O(x^{20}) = \left(\frac{1}{2} - \frac{1}{6}\right)x^{12} + x^{20} + O(x^{20}). \]

Calculating the coefficient:
\[ \frac{1}{2} - \frac{1}{6} = \frac{3}{6} - \frac{1}{6} = \frac{2}{6} = \frac{1}{3}. \]
Thus, we have:
\[ \sin(x^4) - x^4 \cos(x^4) + x^{20} = \frac{1}{3} x^{12} + O(x^{20}). \]

### Step 3: Expand \(e^{2x^4}\)

Using the Taylor series for \(e^x\):
\[ e^{2x^4} = 1 + 2x^4 + \frac{(2x^4)^2}{2!} + O(x^{12}) = 1 + 2x^4 + 2x^8 + O(x^{12}). \]

Thus,
\[ e^{2x^4} - 1 - 2x^4 = 2x^8 + O(x^{12}). \]

### Step 4: Substitute into the limit

Now substituting into the limit:
\[ \lim_{x \to 0} \frac{\frac{1}{3} x^{12} + O(x^{20})}{x^4 (2x^8 + O(x^{12}))} = \lim_{x \to 0} \frac{\frac{1}{3} x^{12}}{2x^{12} + O(x^{16})}. \]

### Step 5: Simplify the limit

This simplifies to:
\[ \lim_{x \to 0} \frac{\frac{1}{3}}{2 + O(x^4)} = \frac{1}{3} \cdot \frac{1}{2} = \frac{1}{6}. \]

### Final Answer

Thus, the limit is:
\[ \boxed{\frac{1}{6}}. \]