We aim to evaluate the limit:

$\lim_{n \to \infty} \frac{\ln(n)}{\sqrt{n}}$

Step-by-step Solution:

1. Examine the growth rates of the numerator and denominator:

   • The numerator $\ln(n)$ grows much slower than any power of $n$.
   • The denominator $\sqrt{n}$ grows much faster than $\ln(n)$ because $\sqrt{n}$ is a power function of $n$.

2. Apply intuition:

   • Since the denominator $\sqrt{n}$ grows much faster than the numerator $\ln(n)$, the fraction $\frac{\ln(n)}{\sqrt{n}}$ should approach 0 as $n \to \infty$.

3. Formal approach using L'Hopital's Rule:

   • This limit is of the indeterminate form $\frac{\infty}{\infty}$. L'Hopital's Rule allows us to differentiate the numerator and denominator: [ \lim_{n \to \infty} \frac{\ln(n)}{\sqrt{n}} = \lim_{n \to \infty} \frac{\frac{d}{dn}[\ln(n)]}{\frac{d}{dn}[\sqrt{n}]}. ]
   • Compute the derivatives: [ \frac{d}{dn}[\ln(n)] = \frac{1}{n}, \quad \frac{d}{dn}[\sqrt{n}] = \frac{1}{2\sqrt{n}}. ]
   • Substituting back: $\lim_{n \to \infty} \frac{\ln(n)}{\sqrt{n}} = \lim_{n \to \infty} \frac{\frac{1}{n}}{\frac{1}{2\sqrt{n}}} = \lim_{n \to \infty} \frac{2\sqrt{n}}{n}.$
   • Simplify the new expression: $\frac{2\sqrt{n}}{n} = \frac{2}{\sqrt{n}}.$

4. Evaluate the simplified limit:

   • As $n \to \infty$, $\sqrt{n} \to \infty$, so: $\lim_{n \to \infty} \frac{2}{\sqrt{n}} = 0.$

Final Answer:

$\lim_{n \to \infty} \frac{\ln(n)}{\sqrt{n}} = 0.$

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Would you like further clarification on any part of this solution?

Related Questions:

1. How does the growth rate of logarithmic functions compare to polynomial functions?
2. What is the general application of L'Hopital's Rule in limits?
3. Why does $\ln(n)$ grow slower than $\sqrt{n}$?
4. How would this change if the denominator were $\ln(n)^k$ instead of $\sqrt{n}$?
5. Can we evaluate the same limit using series approximations?

Tip:

When dealing with limits involving growth rates, always compare the relative growth speeds of the numerator and denominator.