Find lim as x to infinity (ln x)/(sqrt(x-1))
Find $\displaystyle\lim_{x\to\infty}\frac{\ln x}{\sqrt{x-1}}$.
Explanation of Method
Even a half power outgrows the logarithm.
Step 1: The form
Both parts tend to $+\infty$: the form is $\dfrac{\infty}{\infty}$.
Step 2: Apply l'Hôpital's rule
\[ \frac{d}{dx}\ln x = \frac1x, \qquad \frac{d}{dx}\sqrt{x-1} = \frac{1}{2\sqrt{x-1}}, \]
\[ \lim_{x\to\infty}\frac{\ln x}{\sqrt{x-1}} = \lim_{x\to\infty}\frac{1/x}{1/\bigl(2\sqrt{x-1}\bigr)} = \lim_{x\to\infty}\frac{2\sqrt{x-1}}{x}. \]
Step 3: Evaluate the new limit
Extract $\sqrt{x}$ from the numerator: \[\frac{2\sqrt{x-1}}{x} = \frac{2\sqrt{x}\sqrt{1-\tfrac1x}}{x} = \frac{2\sqrt{1-\tfrac1x}}{\sqrt{x}} \longrightarrow \frac{2\cdot1}{+\infty} = 0.\]
Step 4: Verification
\[x = 10^{2}:\ \frac{4.605}{9.950} = 0.4629; \qquad x = 10^{6}:\ \frac{13.816}{1000.0} = 0.01382 \;\;\checkmark\]
\[x = 10^{12}:\ \frac{27.63}{10^{6}} = 2.8\cdot10^{-5} \;\;\checkmark\]
\[\text{The maximum is near } x \approx 15.4, \text{ where the ratio is about } 0.736 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\lim_{x\to\infty}\frac{\ln x}{\sqrt{x-1}} = 0\,}\]
Remark
Remark. The $-1$ under the root is immaterial: $\sqrt{x-1}\sim\sqrt{x}$, so the limit is the case $\alpha = \tfrac12$ of $\dfrac{\ln x}{x^{\alpha}}\to0$. The numbers show how leisurely the decay is — a million still leaves the ratio above $10^{-2}$.