We need to find the limit of the following expression as \( x \to \infty \):

\[
\lim_{x \to \infty} \frac{2x+1}{\sqrt{x^2+3}}
\]

### Step-by-Step Solution

1. **Rewrite the expression**:
   \[
   \lim_{x \to \infty} \frac{2x+1}{\sqrt{x^2+3}}
   \]
   We notice that both the numerator and the denominator grow with \( x \). To handle this, we divide both the numerator and the denominator by \( x \), the highest degree of \( x \) in the denominator.

2. **Divide by \( x \)**:
   \[
   = \lim_{x \to \infty} \frac{2 + \frac{1}{x}}{\sqrt{1 + \frac{3}{x^2}}}
   \]
   - In the numerator: \( \frac{2x + 1}{x} = 2 + \frac{1}{x} \)
   - In the denominator: \( \frac{\sqrt{x^2 + 3}}{x} = \sqrt{1 + \frac{3}{x^2}} \)

3. **Take the limit as \( x \to \infty \)**:
   - As \( x \to \infty \), \( \frac{1}{x} \to 0 \) and \( \frac{3}{x^2} \to 0 \).
   - The expression simplifies to:
     \[
     \lim_{x \to \infty} \frac{2 + 0}{\sqrt{1 + 0}} = \frac{2}{1} = 2
     \]

### Conclusion:
\[
\lim_{x \to \infty} \frac{2x+1}{\sqrt{x^2+3}} = 2
\]

Would you like further clarification or details on this solution? Here are some related questions to explore:

1. What happens if the numerator grows faster than the denominator?
2. How does dividing by the highest power of \( x \) help simplify limits?
3. What is the limit of \(\frac{3x^2 + x}{x^2 - 2}\) as \( x \to \infty \)?
4. How does L'Hopital's Rule work for limits at infinity?
5. Can this method be applied to trigonometric functions?

**Tip**: When solving limits as \( x \to \infty \), divide by the highest power of \( x \) to simplify the expression effectively.

Explore the detailed solution of finding the limit of (2x + 1)/sqrt(x^2 + 3) as x approaches infinity. This problem involves dividing by the highest power of x and applying the limit at infinity theorem, suitable for high school calculus students.

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