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:
\[
\lim_{{x \to \infty}} \frac{2x + 1}{x^2 + 2} = 0
\]

Would you like further clarification or more details?

### 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.

Learn how to evaluate the limit of the rational function (2x+1)/(x^2+2) as x approaches infinity. This solution provides a step-by-step explanation, covering key calculus concepts such as limits at infinity and polynomial degree comparison. Suitable for students in Grades 11-12.

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