The given expression in the image appears to be:

$\lim_{x \to +\infty} \frac{\ln^2(x)}{x-1}$

Let us evaluate this limit step by step:

Step 1: Analyze the growth of numerator and denominator

• The numerator, $\ln^2(x)$, grows much slower than the denominator, $x-1$, as $x \to +\infty$.
• However, since both grow to infinity, this is an indeterminate form of type $\frac{\infty}{\infty}$.

Step 2: Apply L'Hôpital's Rule

Since the limit is indeterminate ($\infty/\infty$), we apply L'Hôpital's Rule: [ \lim_{x \to +\infty} \frac{\ln^2(x)}{x-1} = \lim_{x \to +\infty} \frac{\frac{d}{dx}[\ln^2(x)]}{\frac{d}{dx}[x-1]} ]

Differentiating the numerator:

$\frac{d}{dx}[\ln^2(x)] = 2\ln(x) \cdot \frac{1}{x} = \frac{2\ln(x)}{x}$

Differentiating the denominator:

$\frac{d}{dx}[x-1] = 1$

Thus, the limit becomes: $\lim_{x \to +\infty} \frac{\ln^2(x)}{x-1} = \lim_{x \to +\infty} \frac{\frac{2\ln(x)}{x}}{1} = \lim_{x \to +\infty} \frac{2\ln(x)}{x}$

Step 3: Simplify and re-evaluate the new limit

Now, evaluate: $\lim_{x \to +\infty} \frac{2\ln(x)}{x}$

As $x \to +\infty$, the numerator $\ln(x)$ grows much slower than the denominator $x$. Hence, the fraction tends to $0$.

Final Answer:

$\lim_{x \to +\infty} \frac{\ln^2(x)}{x-1} = 0$

Would you like a detailed explanation of any step? Here are some questions that build on this topic:

1. What are the common scenarios where L'Hôpital's Rule is applicable?
2. How do logarithmic and polynomial functions compare in terms of growth rates?
3. What are some alternative methods for evaluating limits involving logarithmic functions?
4. How would the result change if the denominator were $x^2 - 1$ instead of $x - 1$?
5. Can you explain why $\ln(x)$ grows slower than $x$ as $x \to +\infty$?

Tip: Always check for indeterminate forms ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying L'Hôpital's Rule.