1. **State the problem:** Find the limits of the function $$f(x) = \frac{\ln(x)}{(x-1)^2}$$ as $$x \to +\infty$$, $$x \to 0^+$$, $$x \to 1^+$$, and $$x \to 1^-$$. 2. **Recall important facts:** - The natural logarithm function $$\ln(x)$$ grows slowly to infinity as $$x \to +\infty$$. - The denominator $$(x-1)^2$$ is always positive except at $$x=1$$ where it is zero. - Limits involving division by zero may tend to infinity or negative infinity depending on numerator behavior. 3. **Limit as $$x \to +\infty$$:** - Numerator: $$\ln(x) \to +\infty$$ slowly. - Denominator: $$(x-1)^2 \to +\infty$$ faster (quadratic growth). - So, $$\lim_{x \to +\infty} \frac{\ln(x)}{(x-1)^2} = \frac{+\infty}{+\infty}$$, but denominator dominates. - Using L'Hôpital's Rule: $$\lim_{x \to +\infty} \frac{\ln(x)}{(x-1)^2} = \lim_{x \to +\infty} \frac{\frac{1}{x}}{2(x-1)} = \lim_{x \to +\infty} \frac{1}{2x(x-1)} = 0$$. 4. **Limit as $$x \to 0^+$$:** - Numerator: $$\ln(x) \to -\infty$$. - Denominator: $$(x-1)^2 \to 1$$ (since $$x \to 0$$, $$x-1 \to -1$$). - So, $$\lim_{x \to 0^+} \frac{\ln(x)}{(x-1)^2} = \frac{-\infty}{1} = -\infty$$. 5. **Limit as $$x \to 1^+$$:** - Numerator: $$\ln(1) = 0$$. - Denominator: $$(x-1)^2 \to 0^+$$. - The form is $$\frac{0}{0}$$, so use L'Hôpital's Rule: $$\lim_{x \to 1^+} \frac{\ln(x)}{(x-1)^2} = \lim_{x \to 1^+} \frac{\frac{1}{x}}{2(x-1)} = \lim_{x \to 1^+} \frac{1}{2x(x-1)}$$. - As $$x \to 1^+$$, denominator $$2x(x-1) \to 0^+$$, so the limit tends to $$+\infty$$. 6. **Limit as $$x \to 1^-$$:** - Similar to above, numerator $$\to 0$$, denominator $$\to 0^+$$. - Using L'Hôpital's Rule again: $$\lim_{x \to 1^-} \frac{\ln(x)}{(x-1)^2} = \lim_{x \to 1^-} \frac{1/x}{2(x-1)} = \lim_{x \to 1^-} \frac{1}{2x(x-1)}$$. - As $$x \to 1^-$$, denominator $$2x(x-1) \to 0^-$$, so the limit tends to $$-\infty$$. **Final answers:** $$\lim_{x \to +\infty} \frac{\ln(x)}{(x-1)^2} = 0$$ $$\lim_{x \to 0^+} \frac{\ln(x)}{(x-1)^2} = -\infty$$ $$\lim_{x \to 1^+} \frac{\ln(x)}{(x-1)^2} = +\infty$$ $$\lim_{x \to 1^-} \frac{\ln(x)}{(x-1)^2} = -\infty$$