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Limits — JEE Maths practice question

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

Question

\displaystyle\lim_{x\to 0}\dfrac{\sin(\sin x)-\sin x}{ax^{5}+bx^{3}+c}=-\dfrac{1}{12}

, then

a=2,\,b=0,\,c=1

a=0,\,b=2,\,c\in\mathbb{R}

a=2,\,b\in\mathbb{R},\,c=0

a\in\mathbb{R},\,b=2,\,c=0

correct

Solution

**Step 1: Determine

c

.** As

x\to 0

, the numerator

\to 0

. For the limit to equal the finite non-zero value

-\frac{1}{12}

, denominator must also

\to 0

. So

c=0

. **Step 2: Use sum-to-product.**

\sin C-\sin D=2\cos\!\left(\dfrac{C+D}{2}\right)\sin\!\left(\dfrac{C-D}{2}\right)

\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\to 0

\cos\!\left(\frac{\sin x+x}{2}\right)\to \cos 0=1

.** **Step 4: Use

\sin u\sim u

for small

u

.**

\sin\!\left(\dfrac{\sin x-x}{2}\right)\sim\dfrac{\sin x-x}{2}

**Step 5: The limit reduces to**

\lim_{x\to 0}\dfrac{\sin x-x}{ax^{5}+bx^{3}}=-\dfrac{1}{12}

**Step 6: Apply L'Hôpital twice.**

\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

x

out of denominator.**

=\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

-\frac{1}{12}

.**

-\dfrac{1}{6b}=-\dfrac{1}{12}\;\Rightarrow\;b=2

**Step 9: Note about

a

.** The

a

disappears in the last step, so

a

is unrestricted (any real).

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