Answer is: option2

Solution:

Rewriting the expression:

\[ \lim_{t \to \infty} \frac{\ln \left(1 + \frac{3}{t} \right)}{\frac{1}{t}}. \]

As \( t \to \infty \):

• \(\ln \left(1 + \frac{3}{t} \right) \to \ln(1) = 0\).
• \(\frac{1}{t} \to 0\).

So, the limit is of the form \( \frac{0}{0} \), allowing us to apply L'Hôpital’s Rule.

Using the derivative of \( \ln(1 + x) \) as \( \frac{1}{1+x} \):

Numerator:

\[ \frac{d}{dt} \ln \left( 1 + \frac{3}{t} \right) = \frac{1}{1 + \frac{3}{t}} \left( -\frac{3}{t^2} \right) = \frac{-3/t^2}{1 + \frac{3}{t}}. \]

Denominator:

\[ \frac{d}{dt} \left( \frac{1}{t} \right) = -\frac{1}{t^2}. \]

Now, applying L'Hôpital’s Rule:

\[ \lim_{t\to\infty} \frac{-3/t^2}{1+3/t} \div -1/t^2. \]

Canceling \( -1/t^2 \) from both numerator and denominator:

\[ \lim_{t\to\infty} \frac{3}{1 + 3/t}. \]

As \( t \to \infty \), \( \frac{3}{t} \to 0 \), so:

\[ \frac{3}{1+0} = 3. \]

Final Answer is 3.