The value of `lim_(xto0)(x sin (sinxx)-sin^(2)x)/(x^(6))` equals

Options

A. (a)`1/6`
B. (b)`1/12`
C. (c)`1/18`
D. (d)`1/24`

Answer

To solve the limit problem \( \lim_{x \to 0} \frac{x \sin(\sin x) - \sin^2 x}{x^6} \), we will follow a systematic approach using Taylor series expansions.

### Step-by-Step Solution:

1. **Expand \( \sin x \) using Taylor series:**
The Taylor series expansion of \( \sin x \) around \( x = 0 \) is:
\[ \sin x = x - \frac{x^3}{6} + \frac{x^5}{120} + O(x^7) \]

2. **Substituting \( \sin x \) into \( \sin(\sin x) \):**
Now we substitute \( \sin x \) into itself:
\[ \sin(\sin x) = \sin\left(x - \frac{x^3}{6} + O(x^5)\right) \]
Using the Taylor series again:
\[ \sin(\sin x) = \left(x - \frac{x^3}{6} + O(x^5)\right) - \frac{1}{6}\left(x - \frac{x^3}{6}\right)^3 + O\left(\left(x - \frac{x^3}{6}\right)^5\right) \]

3. **Calculating \( \sin^2 x \):**
We also need \( \sin^2 x \):
\[ \sin^2 x = \left(x - \frac{x^3}{6} + O(x^5)\right)^2 = x^2 - \frac{x^4}{3} + O(x^6) \]

4. **Combine the expressions:**
Now we need to compute \( x \sin(\sin x) - \sin^2 x \):
\[ x \sin(\sin x) = x\left(x - \frac{x^3}{6} + O(x^5)\right) = x^2 - \frac{x^4}{6} + O(x^6) \]
Therefore,
\[ x \sin(\sin x) - \sin^2 x = \left(x^2 - \frac{x^4}{6} + O(x^6)\right) - \left(x^2 - \frac{x^4}{3} + O(x^6)\right) \]
Simplifying this gives:
\[ x \sin(\sin x) - \sin^2 x = -\frac{x^4}{6} + \frac{x^4}{3} + O(x^6) = \frac{x^4}{6} + O(x^6) \]

5. **Substituting back into the limit:**
Now substitute this back into the limit:
\[ \lim_{x \to 0} \frac{\frac{x^4}{6} + O(x^6)}{x^6} = \lim_{x \to 0} \left(\frac{1}{6} \cdot \frac{x^4}{x^6} + \frac{O(x^6)}{x^6}\right) = \lim_{x \to 0} \left(\frac{1}{6} \cdot \frac{1}{x^2} + O(1)\right) \]
As \( x \to 0 \), \( \frac{1}{x^2} \to \infty \) and \( O(1) \) remains bounded.

6. **Final evaluation:**
The limit diverges to infinity, indicating that the original expression approaches infinity as \( x \) approaches 0.

### Conclusion:
Thus, the value of the limit is:
\[ \lim_{x \to 0} \frac{x \sin(\sin x) - \sin^2 x}{x^6} = \infty \]