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Limits: Question

JEE Maths question with a full step-by-step solution.

Question
If lim⁡x→0sin⁡(sin⁡x)−sin⁡xax5+bx3+c=−112\displaystyle\lim_{x\to 0}\dfrac{\sin(\sin x)-\sin x}{ax^{5}+bx^{3}+c}=-\dfrac{1}{12}, then
Aa=2, b=0, c=1a=2,\,b=0,\,c=1
Ba=0, b=2, c∈Ra=0,\,b=2,\,c\in\mathbb{R}
Ca=2, b∈R, c=0a=2,\,b\in\mathbb{R},\,c=0
Da∈R, b=2, c=0a\in\mathbb{R},\,b=2,\,c=0correct
Solution
**Step 1: Determine cc.** As x→0x\to 0, the numerator →0\to 0. For the limit to equal the finite non-zero value −112-\frac{1}{12}, denominator must also →0\to 0. So c=0c=0. **Step 2: Use sum-to-product.**
sin⁡C−sin⁡D=2cos⁡ ⁣(C+D2)sin⁡ ⁣(C−D2)\sin C-\sin D=2\cos\!\left(\dfrac{C+D}{2}\right)\sin\!\left(\dfrac{C-D}{2}\right)
sin⁡(sin⁡x)−sin⁡x=2cos⁡ ⁣(sin⁡x+x2)sin⁡ ⁣(sin⁡x−x2)\sin(\sin x)-\sin x=2\cos\!\left(\dfrac{\sin x+x}{2}\right)\sin\!\left(\dfrac{\sin x-x}{2}\right)
 **Step 3: As x→0x\to 0, cos⁡ ⁣(sin⁡x+x2)→cos⁡0=1\cos\!\left(\frac{\sin x+x}{2}\right)\to \cos 0=1.** **Step 4: Use sin⁡u∼u\sin u\sim u for small uu.**
sin⁡ ⁣(sin⁡x−x2)∼sin⁡x−x2\sin\!\left(\dfrac{\sin x-x}{2}\right)\sim\dfrac{\sin x-x}{2}
 **Step 5: The limit reduces to**
lim⁡x→0sin⁡x−xax5+bx3=−112\lim_{x\to 0}\dfrac{\sin x-x}{ax^{5}+bx^{3}}=-\dfrac{1}{12}
 **Step 6: Apply L'Hôpital twice.**
lim⁡x→0cos⁡x−15ax4+3bx2  →L’H  lim⁡x→0−sin⁡x20ax3+6bx\lim_{x\to 0}\dfrac{\cos x-1}{5ax^{4}+3bx^{2}}\;\xrightarrow{\text{L'H}}\;\lim_{x\to 0}\dfrac{-\sin x}{20ax^{3}+6bx}
 **Step 7: Factor xx out of denominator.**
=lim⁡x→0−sin⁡xx⋅120ax2+6b=−1⋅10+6b=−16b=\lim_{x\to 0}\dfrac{-\sin x}{x}\cdot\dfrac{1}{20ax^{2}+6b}=-1\cdot\dfrac{1}{0+6b}=-\dfrac{1}{6b}
 **Step 8: Set equal to −112-\frac{1}{12}.**
−16b=−112  ⇒  b=2-\dfrac{1}{6b}=-\dfrac{1}{12}\;\Rightarrow\;b=2
 **Step 9: Note about aa.** The aa disappears in the last step, so aa is unrestricted (any real).
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