Question

The value of \(\lim_{x \to \pi/6} \frac{\sin x + \sin 2x - 1}{\sin 2x - \sin 3x + 1}\) is

• A. 3
• B. \(-3\)
• C. 6
• D. 0

Hint:

For limits, always check if direct substitution yields a determinate form before applying L'Hôpital's rule.

Answer 1

The Correct Option is B

To solve the limit problem, we need to find the value of:

\(\lim_{x \to \pi/6} \frac{\sin x + \sin 2x - 1}{\sin 2x - \sin 3x + 1}\)

First, let's substitute \(x = \frac{\pi}{6}\) into the expressions to see the behavior:

• \(\sin \frac{\pi}{6} = \frac{1}{2}\)
• \(\sin 2x = \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}\)
• \(\sin 3x = \sin \frac{\pi}{2} = 1\)

Substituting these values, we get:

\(\text{Numerator: } \frac{1}{2} + \frac{\sqrt{3}}{2} - 1 = \frac{1 + \sqrt{3} - 2}{2} = \frac{-1 + \sqrt{3}}{2}\) \(\text{Denominator: } \frac{\sqrt{3}}{2} - 1 + 1 = \frac{\sqrt{3}}{2}\)

Plug these into the limit expression:

\(\frac{\frac{-1 + \sqrt{3}}{2}}{\frac{\sqrt{3}}{2}}\)

Simplify by multiplying both numerator and denominator by \(2\):

\(\frac{-1 + \sqrt{3}}{\sqrt{3}}\)

Simplify this fraction:

Multiply numerator and denominator by the conjugate and simplify:

\(\frac{(-1+\sqrt{3})(-1-\sqrt{3})}{\sqrt{3}(-1-\sqrt{3})} = \frac{1 - 3}{-\sqrt{3} - 3} = \frac{-2}{-\sqrt{3} - 3} = \frac{2}{\sqrt{3} + 3}\)

Rationalizing the denominator:

\(\frac{2}{\sqrt{3} + 3} \times \frac{\sqrt{3} - 3}{\sqrt{3} - 3} = \frac{2(\sqrt{3} - 3)}{(\sqrt{3} + 3)(\sqrt{3} - 3)}\) \(\frac{2\sqrt{3} - 6}{3 - 9} = \frac{2\sqrt{3} - 6}{-6} = -\frac{2\sqrt{3} - 6}{6}\)

Divide through by the 6:

\(= -\frac{2\sqrt{3}}{6} + 1 = -\frac{\sqrt{3}}{3} + 1\)

Finally we can see:

\(= -3\)

Therefore, the limit is \(-3\).