Question

Evaluate \[ \lim_{x\to \frac{\pi}{6}} \frac{3\sin x-\sqrt{3}\cos x}{6x-\pi} \]

• A. \(-\dfrac{1}{\sqrt{3}}\)
• B. \(\dfrac{1}{\sqrt{3}}\)
• C. \(\dfrac{1}{\sqrt{2}}\)
• D. \(-\dfrac{1}{\sqrt{2}}\)

Hint:

Whenever a limit gives the indeterminate form \[ \frac{0}{0}, \] L'Hospital's Rule can be applied by differentiating the numerator and denominator separately.

Answer 1

The Correct Option is B

Step 1: Substitute the limiting value.
Substituting \[ x=\frac{\pi}{6}, \] we get Numerator: \[ 3\sin\frac{\pi}{6}-\sqrt{3}\cos\frac{\pi}{6} \] Using \[ \sin\frac{\pi}{6}=\frac{1}{2} \] and \[ \cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}, \] we obtain \[ 3\left(\frac{1}{2}\right)-\sqrt{3}\left(\frac{\sqrt{3}}{2}\right) \] \[ =\frac{3}{2}-\frac{3}{2}=0 \] Denominator: \[ 6\left(\frac{\pi}{6}\right)-\pi \] \[ =\pi-\pi=0 \] Thus, the limit is of the indeterminate form \[ \frac{0}{0} \]

Step 2: Apply L'Hospital's Rule.
Differentiate numerator and denominator separately. Derivative of numerator: \[ \frac{d}{dx}(3\sin x-\sqrt{3}\cos x) \] \[ =3\cos x+\sqrt{3}\sin x \] Derivative of denominator: \[ \frac{d}{dx}(6x-\pi)=6 \] Therefore, \[ \lim_{x\to \frac{\pi}{6}} \frac{3\sin x-\sqrt{3}\cos x}{6x-\pi} = \lim_{x\to \frac{\pi}{6}} \frac{3\cos x+\sqrt{3}\sin x}{6} \]

Step 3: Substitute the value again.
Substituting \[ x=\frac{\pi}{6}, \] we get \[ \frac{ 3\cos\frac{\pi}{6} +\sqrt{3}\sin\frac{\pi}{6} }{6} \] Using \[ \cos\frac{\pi}{6}=\frac{\sqrt{3}}{2} \] and \[ \sin\frac{\pi}{6}=\frac{1}{2}, \] we obtain \[ \frac{ 3\left(\frac{\sqrt{3}}{2}\right) +\sqrt{3}\left(\frac{1}{2}\right) }{6} \] \[ = \frac{ \frac{3\sqrt{3}}{2} +\frac{\sqrt{3}}{2} }{6} \] \[ = \frac{ \frac{4\sqrt{3}}{2} }{6} \] \[ = \frac{2\sqrt{3}}{6} \] \[ = \frac{\sqrt{3}}{3} \] \[ = \frac{1}{\sqrt{3}} \]

Step 4: Final conclusion.
Hence, \[ \boxed{\frac{1}{\sqrt{3}}} \]