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Limits — JEE Maths practice question

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

Question
lim⁡x→0ex−esin⁡x2(x−sin⁡x)= ?\displaystyle\lim_{x\to 0}\dfrac{e^{x}-e^{\sin x}}{2(x-\sin x)}=\,?
A−12-\dfrac{1}{2}
B11
C32\dfrac{3}{2}
D12\dfrac{1}{2}correct
Solution
Step 1: Factor esin⁡xe^{\sin x} from the numerator:
ex−esin⁡x=esin⁡x(ex−sin⁡x−1)e^{x}-e^{\sin x}=e^{\sin x}\left(e^{x-\sin x}-1\right)
 Step 2: Substitute back:
lim⁡x→0esin⁡x(ex−sin⁡x−1)2(x−sin⁡x)\lim_{x\to 0}\dfrac{e^{\sin x}\left(e^{x-\sin x}-1\right)}{2(x-\sin x)}
 Step 3: Split into a product of three factors:
=lim⁡x→0esin⁡x⋅ex−sin⁡x−1x−sin⁡x⋅12=\lim_{x\to 0}e^{\sin x}\cdot\dfrac{e^{x-\sin x}-1}{x-\sin x}\cdot\dfrac{1}{2}
 Step 4: Evaluate each factor. esin⁡x→e0=1e^{\sin x}\to e^{0}=1. Let u=x−sin⁡xu=x-\sin x. Then u→0u\to 0 and eu−1u→1\dfrac{e^{u}-1}{u}\to 1. Step 5: Multiply:
1⋅1⋅12=121\cdot 1\cdot\dfrac{1}{2}=\dfrac{1}{2}
 Correct answer: (4) 12\dfrac{1}{2}
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