Question:

The value of \( \lim_{x \to 0} [ \dfrac{x - \sin 2x}{x - \sin 5x} ] \) (rounded off to two decimal places) is \(\underline{\hspace{1cm}}\). 
 

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For limits involving \( \sin(kx) \), use the small-angle expansion to simplify.
Updated On: Jan 2, 2026
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Correct Answer: 0.25

Solution and Explanation

Using the approximation for small \( x \):
\[ \sin kx \approx kx - \frac{(kx)^3}{6} \]
Thus,
\[ x - \sin 2x \approx x - \left(2x - \frac{8x^3}{6}\right) = -x + \frac{4x^3}{3} \]
\[ x - \sin 5x \approx x - \left(5x - \frac{125x^3}{6}\right) = -4x + \frac{125x^3}{6} \]
Taking the limit as \( x \to 0 \):
\[ \lim_{x \to 0} \frac{x - \sin 2x}{x - \sin 5x} = \frac{-1}{-4} = \frac{1}{4} = 0.25 \]
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