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

Step 1: Test the form at $x=\dfrac{\pi}{3}$.
Top: $\tan^3\dfrac{\pi}{3}-3\tan\dfrac{\pi}{3}=(\sqrt3)^3-3\sqrt3=3\sqrt3-3\sqrt3=0$. Bottom: $\cos\left(\dfrac{\pi}{3}+\dfrac{\pi}{6}\right)=\cos\dfrac{\pi}{2}=0$. So it is $\dfrac00$.
Step 2: Decide to use L'Hospital's rule.
Because of the $\dfrac00$ form, differentiate top and bottom.
Step 3: Differentiate the top.
\[ \frac{d}{dx}(\tan^3x-3\tan x)=3\tan^2x\sec^2x-3\sec^2x=3\sec^2x(\tan^2x-1). \]
Step 4: Differentiate the bottom.
\[ \frac{d}{dx}\cos\left(x+\frac{\pi}{6}\right)=-\sin\left(x+\frac{\pi}{6}\right). \]
Step 5: Put $x=\dfrac{\pi}{3}$ into the needed values.
$\tan^2\dfrac{\pi}{3}=3$, $\sec^2\dfrac{\pi}{3}=4$, and $\sin\left(\dfrac{\pi}{3}+\dfrac{\pi}{6}\right)=\sin\dfrac{\pi}{2}=1$.
Step 6: Combine.
\[ \frac{3(4)(3-1)}{-1}=\frac{24}{-1}=-24. \] \[ \boxed{-24} \]