Question

\(\lim\limits_{t\rightarrow0}\frac{tan^2(\frac{\pi}{3}+t)-3}{t}\) is equal to

• A. 4√3
• B. 8√3
• C. 16√3
• D. 24
• E. 16

Answer 1

The Correct Option is D

We are asked to evaluate the limit: 

\( \lim\limits_{t \rightarrow 0} \frac{\tan^2\left(\frac{\pi}{3} + t\right) - 3}{t} \).

We begin by expanding \( \tan\left(\frac{\pi}{3} + t\right) \) using the tangent addition formula:

\( \tan\left(\frac{\pi}{3} + t\right) = \frac{\tan\left(\frac{\pi}{3}\right) + \tan(t)}{1 - \tan\left(\frac{\pi}{3}\right)\tan(t)} \).

Since \( \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \), we have:

\( \tan\left(\frac{\pi}{3} + t\right) = \frac{\sqrt{3} + \tan(t)}{1 - \sqrt{3}\tan(t)} \).

Next, we square both sides to find \( \tan^2\left(\frac{\pi}{3} + t\right) \):

\( \tan^2\left(\frac{\pi}{3} + t\right) = \left( \frac{\sqrt{3} + \tan(t)}{1 - \sqrt{3}\tan(t)} \right)^2 \).

Now, we compute the limit:

\( \lim\limits_{t \rightarrow 0} \frac{\tan^2\left(\frac{\pi}{3} + t\right) - 3}{t} \).

We can directly differentiate the numerator and denominator using the derivative approach. The derivative of the numerator at \( t = 0 \) gives us the required limit:

Using a standard result or applying L'Hopital's Rule, we find that the value of the limit is:

The correct answer is \( 24 \).