Question

Evaluate: \[ \lim_{x\to \frac{\pi}{3}} \frac{\tan^3x-3\tan x} {\cos\left(x+\frac{\pi}{6}\right)} \]

• A. \(12\)
• B. \(24\)
• C. \(-24\)
• D. \(-12\)

Hint:

For trigonometric limits producing \(\frac00\):

• first check identities,

• then apply L'Hospital's Rule if necessary.
Useful values: \[ \tan\frac{\pi}{3}=\sqrt3, \qquad \sec\frac{\pi}{3}=2 \]

Answer 1

The Correct Option is C

Concept: Trigonometric limits often require:

• factorization,

• standard identities,

• or L'Hospital's Rule.
We simplify carefully before evaluating the limit.

Step 1: Substitute \(x=\frac{\pi}{3}\).
Since: \[ \tan\frac{\pi}{3}=\sqrt3 \] the numerator becomes: \[ (\sqrt3)^3-3(\sqrt3) = 3\sqrt3-3\sqrt3 = 0 \] The denominator becomes: \[ \cos\left(\frac{\pi}{3}+\frac{\pi}{6}\right) = \cos\frac{\pi}{2} = 0 \] Hence we get the indeterminate form: \[ \frac00 \] Therefore, apply L'Hospital's Rule.

Step 2: Differentiate numerator and denominator.
Differentiate numerator: \[ \frac{d}{dx}(\tan^3x-3\tan x) = 3\tan^2x\sec^2x-3\sec^2x \] \[ = 3\sec^2x(\tan^2x-1) \] Differentiate denominator: \[ \frac{d}{dx}\cos\left(x+\frac{\pi}{6}\right) = -\sin\left(x+\frac{\pi}{6}\right) \] Thus: \[ \lim_{x\to \frac{\pi}{3}} \frac{\tan^3x-3\tan x} {\cos\left(x+\frac{\pi}{6}\right)} = \lim_{x\to \frac{\pi}{3}} \frac{3\sec^2x(\tan^2x-1)} {-\sin\left(x+\frac{\pi}{6}\right)} \]

Step 3: Evaluate at \(x=\frac{\pi}{3}\).
We know: \[ \tan\frac{\pi}{3}=\sqrt3 \quad \Rightarrow \quad \tan^2\frac{\pi}{3}=3 \] and \[ \sec\frac{\pi}{3}=2 \quad \Rightarrow \quad \sec^2\frac{\pi}{3}=4 \] Also, \[ \sin\left(\frac{\pi}{3}+\frac{\pi}{6}\right) = \sin\frac{\pi}{2} = 1 \] Substituting: \[ = \frac{3(4)(3-1)}{-1} \] \[ = \frac{24}{-1} = -24 \] Hence, \[ \boxed{-24} \]