The given limit is:

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

Step-by-Step Solution:

1. Examine the degrees of the terms:

   • The degree of the numerator, $2x + 1$, is 1 (since $2x$ is the dominant term).
   • The degree of the denominator, $x^2 + 2$, is 2 (since $x^2$ is the dominant term).

2. Divide both numerator and denominator by the highest degree of $x$ in the denominator (which is $x^2$):

$\frac{2x + 1}{x^2 + 2} = \frac{\frac{2x + 1}{x^2}}{\frac{x^2 + 2}{x^2}} = \frac{\frac{2}{x} + \frac{1}{x^2}}{1 + \frac{2}{x^2}}$

3. Evaluate the limit as $x \to \infty$:

   • As $x \to \infty$, both $\frac{2}{x} \to 0$ and $\frac{1}{x^2} \to 0$.
   • Also, $\frac{2}{x^2} \to 0$.

   Therefore, the expression simplifies to:

$\lim_{{x \to \infty}} \frac{0 + 0}{1 + 0} = 0$

Conclusion:

$\lim_{{x \to \infty}} \frac{2x + 1}{x^2 + 2} = 0$

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Related Questions:

1. What is the general method for finding limits at infinity?
2. How do you determine the dominant terms in a rational function?
3. What are other ways to simplify limits without dividing by the highest power of $x$?
4. How do limits at infinity differ from limits approaching a finite value?
5. Can L'Hopital's Rule be used here to confirm the result?

Tip:

When analyzing limits involving rational functions, comparing the degrees of the numerator and denominator can quickly reveal the limit's behavior at infinity.