Question

Evaluate \( \lim_{x \to 0} \frac{\sin 2x + \sin 5x}{\sin 4x + \sin 6x} \):

• A. \( \frac{2}{5} \)
• B. \( \frac{7}{5} \)
• C. \( \frac{3}{7} \)
• D. \( \frac{7}{10} \)
• E. \( \frac{5}{7} \)

Hint:

When the limit leads to an indeterminate form \( \frac{0}{0} \), apply L'Hopital's rule by differentiating the numerator and denominator.

Answer 1

The Correct Option is D

We can apply L'Hopital's rule because the expression evaluates to \( \frac{0}{0} \) when \( x = 0 \). 
Differentiate the numerator and denominator: \[ {Numerator:} \quad \frac{d}{dx} \left( \sin 2x + \sin 5x \right) = 2\cos 2x + 5\cos 5x \] \[ {Denominator:} \quad \frac{d}{dx} \left( \sin 4x + \sin 6x \right) = 4\cos 4x + 6\cos 6x \] Now, substitute \( x = 0 \) into these derivatives: \[ {Numerator at } x = 0: \quad 2\cos 0 + 5\cos 0 = 2 + 5 = 7 \] \[ {Denominator at } x = 0: \quad 4\cos 0 + 6\cos 0 = 4 + 6 = 10 \] Thus, the limit is: \[ \frac{7}{10} \]