Using the series expansion of \( \ln(1+x) \) in a limit:
$$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \dots$$
Substitute into the limit:
$$\lim_{x \to 0} \frac{\ln(1+x)}{x} = \lim_{x \to 0} \frac{x - \frac{x^2}{2} + \frac{x^3}{3} + \dots}{x}$$
$$= \lim_{x \to 0} \left(1 - \frac{x}{2} + \frac{x^2}{3} + \dots \right) = 1$$
$$\text{ANSWER: } \lim_{x \to 0} \frac{\ln(1+x)}{x} = 1$$
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