Question

\(\lim\limits_{t\rightarrow0}\frac{\sin2t}{8t^2+4t}\) is equal to

• A. \(\frac{1}{2}\)
• B. \(\frac{2}{5}\)
• C. \(\frac{1}{6}\)
• D. \(\frac{1}{3}\)
• E. 1

Answer 1

The Correct Option is A

We are given the limit:

\[ \lim_{t \to 0} \frac{\sin(2t)}{8t^2 + 4t} \]

We can simplify the expression:

\[ \frac{\sin(2t)}{8t^2 + 4t} = \frac{\sin(2t)}{4t(2t + 1)} \]

As \( t \to 0 \), \( \sin(2t) \approx 2t \), so:

\[ \frac{\sin(2t)}{4t(2t + 1)} \approx \frac{2t}{4t(2t + 1)} = \frac{2}{4(2t + 1)} \]

Now, taking the limit as \( t \to 0 \), we get:

\[ \lim_{t \to 0} \frac{2}{4(2t + 1)} = \frac{2}{4(0 + 1)} = \frac{2}{4} = \frac{1}{2} \]

Answer: \( \frac{1}{2} \)