Question

If \[ \lim_{x \to 0} \frac{\sin(\sin x) - \sin x}{ax^3 + bx^5 + c} = -\frac{1}{12}, \] then

• A. \(a=2, b\in \mathbb{R}, c=0\)
• B. \(a=-2, b\in \mathbb{R}, c=0\)
• C. \(a=1, b\in \mathbb{R}, c=0\)
• D. \(a=-1, b\in \mathbb{R}, c=0\)

Hint:

Always expand inner function first, then outer function carefully—sign mistakes are common.

Answer 1

The Correct Option is B

Concept: Use series expansion up to \(x^3\) terms.

Step 1: Expand \(\sin x\). \[ \sin x = x - \frac{x^3}{6} + O(x^5) \]

Step 2: Expand \(\sin(\sin x)\). Let \(u = \sin x\), then: \[ \sin u = u - \frac{u^3}{6} + O(u^5) \] \[ \Rightarrow \sin(\sin x) = \left(x - \frac{x^3}{6}\right) - \frac{1}{6}\left(x^3\right) + O(x^5) \] \[ = x - \frac{x^3}{6} - \frac{x^3}{6} + O(x^5) = x - \frac{x^3}{3} + O(x^5) \]

Step 3: Numerator. \[ \sin(\sin x) - \sin x = \left(x - \frac{x^3}{3}\right) - \left(x - \frac{x^3}{6}\right) \] \[ = -\frac{x^3}{6} + O(x^5) \]

Step 4: Denominator condition. For limit to exist: \[ c = 0 \] \[ \Rightarrow \frac{-\frac{1}{6}x^3}{ax^3} = -\frac{1}{6a} \]

Step 5: Compare. \[ -\frac{1}{6a} = -\frac{1}{12} \Rightarrow a = -2 \]

Step 6: Conclusion. \[ {a=-2,\; b\in \mathbb{R},\; c=0} \]