Find lim as x to infinity (ln x)/(x^4)
Find $\displaystyle\lim_{x\to\infty}\frac{\ln x}{x^{4}}$.
Explanation of Method
One application of the rule settles it.
Step 1: The form
Both parts tend to $+\infty$, so the form is $\dfrac{\infty}{\infty}$.
Step 2: Apply l'Hôpital's rule
\[ \lim_{x\to\infty}\frac{\ln x}{x^{4}} = \lim_{x\to\infty}\frac{1/x}{4x^{3}} = \lim_{x\to\infty}\frac{1}{4x^{4}}. \]
One step suffices: differentiating the logarithm produces a negative power, so the new denominator grows even faster than the old one.
Step 3: Verification
\[x = 10:\ \frac{2.3026}{10^{4}} = 2.3\cdot10^{-4}; \qquad x = 100:\ \frac{4.6052}{10^{8}} = 4.6\cdot10^{-8} \;\;\checkmark\]
\[\text{Substituting } x = e^{t}: \ \frac{t}{e^{4t}} \to 0 \text{ by Problem 378 with } n = 1,\ a = 4 \;\;\checkmark\]
\[\text{The maximum of } \frac{\ln x}{x^{4}} \text{ is at } x = e^{1/4}, \text{ where it equals } \frac{1}{4e} \approx 0.092 \;\;\checkmark\]
Result
Final answer:
\[\boxed{\,\lim_{x\to\infty}\frac{\ln x}{x^{4}} = 0\,}\]
Remark
Remark. The same argument gives $\lim\dfrac{\ln x}{x^{\alpha}} = 0$ for every $\alpha>0$: the logarithm grows more slowly than any positive power, however small the exponent. The substitution $x = e^{t}$ in the verification shows this is Problem 378 in different clothing.