We are tasked with evaluating the limit:

$\lim_{{x \to \infty}} \frac{2x^2 + 3x + 5}{x^2 + 1}$

Step 1: Analyze the degree of the numerator and denominator

• The degree of the numerator, $2x^2 + 3x + 5$, is 2 (since the highest power of $x$ is $x^2$).
• The degree of the denominator, $x^2 + 1$, is also 2 (since the highest power of $x$ is $x^2$).

When the degrees of the numerator and denominator are the same, the limit as $x$ approaches infinity is given by the ratio of the leading coefficients of $x^2$.

Step 2: Identify the leading coefficients

• The leading coefficient of the numerator $2x^2$ is 2.
• The leading coefficient of the denominator $x^2$ is 1.

Step 3: Take the limit

As $x \to \infty$, the lower-order terms (like $3x$, $5$, and $1$) become negligible compared to the highest-degree terms. Thus, we focus on the ratio of the leading terms:

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

Final Answer:

$\lim_{{x \to \infty}} \frac{2x^2 + 3x + 5}{x^2 + 1} = 2$

Would you like any further details or clarifications on this solution?

Here are 5 related questions to extend your understanding:

1. What happens if the degree of the numerator is higher than the denominator?
2. How do we handle limits where the degree of the denominator is higher than the numerator?
3. How would the result change if the coefficients of the highest powers were negative?
4. Can you generalize the result for any rational function where the degrees of the numerator and denominator are the same?
5. How would the solution change if we were taking the limit as $x \to -\infty$?

Tip: When dealing with limits at infinity for rational functions, the degrees of the numerator and denominator often provide key insights into the behavior of the function.