1. The problem asks us to find the limit \( \lim_{n \to +\infty} \frac{\ln(n)}{\sqrt{n+1}} \). 2. As \( n \to +\infty \), both \( \ln(n) \) and \( \sqrt{n+1} \) grow without bound, but we need to compare their growth rates. 3. Recall that the logarithm function grows slower than any positive power of \( n \), so \( \ln(n) \) grows slower than \( \sqrt{n} = n^{1/2} \). 4. To make this precise, let's use the fact that for large \( n \), \( \sqrt{n+1} \approx \sqrt{n} = n^{1/2} \). 5. Consider the expression $$ \frac{\ln(n)}{\sqrt{n+1}} \approx \frac{\ln(n)}{n^{1/2}}. $$ 6. Applying L'Hôpital's rule or using standard limits, the limit of \( \frac{\ln(n)}{n^{1/2}} \) as \( n \to \infty \) is 0 because the denominator grows faster than the numerator. 7. Hence, $$ \lim_{n \to +\infty} \frac{\ln(n)}{\sqrt{n+1}} = 0. $$ Therefore, the final answer is \( 0 \).