State the indeterminate form of the limit of ln x / sqrt(x) as x approaches infinity, then use L Hopital rule to find the limit.

Step-by-step solution

1. 1

   Identify the indeterminate form. As $x\to+\infty$ both $\ln x \to +\infty$ and $\sqrt{x} \to +\infty$, so the quotient has the form $\dfrac{\infty}{\infty}$. That is one of the two forms L'Hopital's Rule handles directly.

2. 2

   Differentiate numerator and denominator separately. L'Hopital's Rule replaces $\dfrac{f}{g}$ by $\dfrac{f'}{g'}$ — this is not the quotient rule, and mixing the two is the classic error:

   $f(x) = \ln x \Rightarrow f'(x) = \frac{1}{x}, \qquad g(x) = x^{1/2} \Rightarrow g'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$

3. 3

   Form and simplify the new quotient.

   $\frac{f'(x)}{g'(x)} = \frac{1/x}{1/(2\sqrt{x})} = \frac{2\sqrt{x}}{x} = \frac{2}{\sqrt{x}}$

   The simplification is what makes the rule pay off: the new expression is no longer indeterminate.

4. 4

   Take the limit.

   $\lim_{x\to+\infty}\frac{2}{\sqrt{x}} = 0$

   so the original limit is $0$ as well. One application of the rule was enough.

5. 5

   Interpret the result. This is the standard statement that logarithms grow more slowly than any positive power of $x$: $\ln x = o(x^{p})$ for every $p > 0$. Here $p = \tfrac12$, and the same argument works for any $p$.

6. 6

   Verify numerically. At $x = 10^4$: $\dfrac{\ln 10^4}{\sqrt{10^4}} = \dfrac{9.2103}{100} = 0.092103$. At $x = 10^8$: $\dfrac{18.4207}{10^4} = 0.00184207$. At $x = 10^{16}$: $\dfrac{36.8414}{10^8} = 3.68\times10^{-7}$ ✓ — decreasing toward $0$, and a symbolic limit returns exactly $0$.