The problem asks to evaluate the limit:

$\lim_{x \to \infty} \frac{2x^2}{x^2 + 1}$

Step-by-Step Solution:

1. Identify the highest degree terms: The degrees of both the numerator and denominator are the same (both are $x^2$). This is crucial for determining how the limit behaves as $x$ approaches infinity.

2. Factor out the highest power of $x$ from both the numerator and the denominator: $\frac{2x^2}{x^2 + 1} = \frac{x^2(2)}{x^2(1 + \frac{1}{x^2})}$

3. Simplify the expression: $\frac{x^2(2)}{x^2(1 + \frac{1}{x^2})} = \frac{2}{1 + \frac{1}{x^2}}$

4. Take the limit as $x \to \infty$: As $x$ goes to infinity, $\frac{1}{x^2} \to 0$. Therefore, the expression simplifies to: $\frac{2}{1 + 0} = 2$

Final Answer:

$\lim_{x \to \infty} \frac{2x^2}{x^2 + 1} = 2$

Would you like a more detailed explanation or have any further questions?

Here are 5 related questions to expand on this:

1. How would you evaluate $\lim_{x \to \infty} \frac{3x^3}{x^3 + 2x}$?
2. What happens if the degree of the numerator is higher than the denominator?
3. How would the limit change if the degrees of the numerator and denominator were different?
4. Can you apply L'Hopital's Rule to this limit problem? Why or why not?
5. What happens if you substitute $x = 0$ in the given function?

Tip: When dealing with rational functions and limits at infinity, always check the degrees of the highest power terms in both the numerator and denominator. This helps determine the limit behavior efficiently!