To solve the given limit using L'Hôpital's rule:

We are tasked to compute:

$\lim_{x \to \infty} \frac{\ln\left(1 + \frac{2}{x}\right)}{\pi - 2\arctan(x)}.$

Step 1: Analyze the behavior of the numerator and denominator

• Numerator: As $x \to \infty$, $\frac{2}{x} \to 0$, so $\ln\left(1 + \frac{2}{x}\right) \to \ln(1) = 0$.
• Denominator: As $x \to \infty$, $\arctan(x) \to \frac{\pi}{2}$, so $\pi - 2\arctan(x) \to \pi - \pi = 0$.

Thus, the limit is in an indeterminate form $\frac{0}{0}$, and we can apply L'Hôpital's rule.

Step 2: Differentiate numerator and denominator

• Derivative of the numerator: $\frac{d}{dx} \ln\left(1 + \frac{2}{x}\right) = \frac{1}{1 + \frac{2}{x}} \cdot \left(-\frac{2}{x^2}\right) = \frac{-\frac{2}{x^2}}{1 + \frac{2}{x}}.$

• Derivative of the denominator: $\frac{d}{dx} \left(\pi - 2\arctan(x)\right) = -2 \cdot \frac{1}{1 + x^2}.$

Step 3: Rewrite the new fraction

The limit becomes: $\lim_{x \to \infty} \frac{\frac{-\frac{2}{x^2}}{1 + \frac{2}{x}}}{-\frac{2}{1 + x^2}}.$

Simplify: $= \lim_{x \to \infty} \frac{\frac{2}{x^2(1 + \frac{2}{x})}}{\frac{2}{1 + x^2}}.$

Cancel the factor of 2: $= \lim_{x \to \infty} \frac{\frac{1}{x^2(1 + \frac{2}{x})}}{\frac{1}{1 + x^2}}.$

Simplify further: $= \lim_{x \to \infty} \frac{1 + x^2}{x^2\left(1 + \frac{2}{x}\right)}.$

Step 4: Evaluate the limit as $x \to \infty$

As $x \to \infty$:

• $1 + x^2 \to x^2$,
• $1 + \frac{2}{x} \to 1$.

Thus: $\lim_{x \to \infty} \frac{1 + x^2}{x^2\left(1 + \frac{2}{x}\right)} = \lim_{x \to \infty} \frac{x^2}{x^2} = 1.$

Final Answer:

$\boxed{1}.$

Would you like me to clarify any step or go deeper into the explanation? Here are 5 related questions for further exploration:

1. What is the general procedure to apply L'Hôpital's rule to indeterminate forms?
2. How do the properties of logarithmic functions help simplify limits?
3. Can you extend this problem to a similar one involving $\arcsin(x)$ or $\arccos(x)$?
4. What happens if the numerator and denominator approach infinity instead of zero?
5. How does $\arctan(x)$ behave as $x \to -\infty$?

Tip: Always check whether a limit is indeterminate before applying L’Hôpital’s rule to ensure it is valid!