Question

Find the limit:

$\lim_{x \to 0} \frac{ln(1+3x)}{x}$

Answer 1

3

Answer 2

The limit is of the $\frac{0}{0}$ indeterminate form. Using the standard limit $\lim_{u \to 0} \frac{\ln(1+u)}{u} = 1$, we manipulate the expression:

$\lim_{x \to 0} \frac{\ln(1+3x)}{x} = \lim_{x \to 0} \frac{\ln(1+3x)}{3x} \cdot 3$.

Let $u = 3x$. As $x \to 0$, $u \to 0$.

The limit becomes $3 \cdot \lim_{u \to 0} \frac{\ln(1+u)}{u} = 3 \cdot 1 = 3$.