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To find the values of \( a \) and \( b \) in the given equation:

\[
\lim_{x \to \infty} \left( \frac{2x^2 + 1}{x - 3} - ax - b \right) = 0
\]

### Step 1: Analyze the behavior of \( \frac{2x^2 + 1}{x - 3} \) as \( x \to \infty \)

For large \( x \), we approximate the expression:

\[
\frac{2x^2 + 1}{x - 3} \approx \frac{2x^2}{x} = 2x
\]

So as \( x \to \infty \), the leading term of the expression is \( 2x \). The subleading terms can be considered smaller as \( x \to \infty \).

### Step 2: Set the dominant term equal to \( ax \)

Since the limit must hold true as \( x \to \infty \), the \( x \)-terms must cancel out. Therefore, we set:

\[
2x = ax
\]

This gives:

\[
a = 2
\]

### Step 3: Consider the constant term

Now, subtract \( ax \) from the original expression:

\[
\frac{2x^2 + 1}{x - 3} - 2x = \frac{2x^2 + 1 - 2x(x - 3)}{x - 3} = \frac{2x^2 + 1 - 2x^2 + 6x}{x - 3} = \frac{6x + 1}{x - 3}
\]

For large \( x \), we approximate:

\[
\frac{6x + 1}{x - 3} \approx 6
\]

### Step 4: Set the constant term equal to \( b \)

The remaining constant must cancel out, so:

\[
6 = b
\]

### Final Answer:
\[
a = 2, \quad b = 6
\]

Would you like further details on any steps?

### Related Questions:
1. How can we analyze the limit at \( x \to 0 \) for similar expressions?
2. What happens if the denominator has a higher-degree polynomial than the numerator?
3. How does the method change if the expression had a cubic term in the numerator?
4. Can we solve this limit problem using L'Hopital's Rule?
5. How do limits at infinity behave in rational functions?

### Tip:
When solving limits at infinity, always first consider the highest-degree terms in both the numerator and denominator, as they dominate the behavior of the function for large \( x \).

Discover how to solve the limit equation lim(x→∞) [(2x^2 + 1) / (x - 3) - ax - b] = 0 by finding the values of a and b. Learn key concepts such as rational functions and asymptotic behavior. Suitable for high school or early undergraduate students.

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