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: Confirm the 0/0 form at $ x=\pi/6 $.
Numerator: $ 3\cdot\frac{1}{2}-\sqrt{3}\cdot\frac{\sqrt{3}}{2}=\frac{3}{2}-\frac{3}{2}=0 $. Denominator: $ 6\cdot\frac{\pi}{6}-\pi=0 $. Confirmed $ 0/0 $.
Step 2: Substitute $ x=\frac{\pi}{6}+h $, $ h\to 0 $.
Denominator: $ 6h $.
Step 3: Expand the numerator using sum formulas.
\[ 3\sin\!\left(\tfrac{\pi}{6}+h\right) = 3\!\left(\tfrac{1}{2}\cos h+\tfrac{\sqrt{3}}{2}\sin h\right) = \tfrac{3}{2}\cos h+\tfrac{3\sqrt{3}}{2}\sin h \] \[ \sqrt{3}\cos\!\left(\tfrac{\pi}{6}+h\right) = \sqrt{3}\!\left(\tfrac{\sqrt{3}}{2}\cos h-\tfrac{1}{2}\sin h\right) = \tfrac{3}{2}\cos h-\tfrac{\sqrt{3}}{2}\sin h \]
Step 4: Subtract and simplify.
Numerator $ = \tfrac{3\sqrt{3}}{2}\sin h+\tfrac{\sqrt{3}}{2}\sin h = 2\sqrt{3}\sin h $.
Step 5: Form the limit.
\[ \lim_{h\to 0}\frac{2\sqrt{3}\sin h}{6h} = \frac{2\sqrt{3}}{6}\cdot 1 = \frac{\sqrt{3}}{3} \]
Step 6: Simplify to the standard form.
\[ \frac{\sqrt{3}}{3} = \frac{1}{\sqrt{3}} \] \[ \boxed{\dfrac{1}{\sqrt{3}}} \]