Limits
Trigonometric Limits
JEE Advanced
Grade 11

Question:

$\lim_{x \to 0} \frac{\sin x - \tan x}{x^3}$

Step-by-Step Solution

Key Concept: $\sin x - \tan x = \sin x \left(1 - \frac{1}{\cos x}\right) = \sin x \cdot \frac{\cos x - 1}{\cos x} \approx x \cdot \frac{-x^2/2}{1} = -\frac{x^3}{2}$. Dividing by $x^3$: limit $= -\frac{1}{2}$.
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Correct Answer: -\frac{1}{2}

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