lim x -> +inf, sqrt(4(x^2) + 3x) - 2x? Do it step by step, using traditional methods

To find the limit of the expression \( \lim_{x \to +\infty} \left( \sqrt{4x^2 + 3x} - 2x \right) \), we can proceed by simplifying the expression step-by-step. Here's how:

### Step 1: Rewrite the Expression
We start with:
\[
\lim_{x \to +\infty} \left( \sqrt{4x^2 + 3x} - 2x \right)
\]
Our goal is to simplify this expression as \( x \) approaches infinity.

### Step 2: Rationalize the Expression
Since the expression involves a square root and subtraction, it helps to multiply and divide by the conjugate:
\[
\lim_{x \to +\infty} \frac{\left( \sqrt{4x^2 + 3x} - 2x \right) \left( \sqrt{4x^2 + 3x} + 2x \right)}{\sqrt{4x^2 + 3x} + 2x}
\]
This simplifies to:
\[
\lim_{x \to +\infty} \frac{(4x^2 + 3x) - (2x)^2}{\sqrt{4x^2 + 3x} + 2x}
\]

### Step 3: Simplify the Numerator
Expanding the terms in the numerator:
\[
(4x^2 + 3x) - 4x^2 = 3x
\]
So the expression becomes:
\[
\lim_{x \to +\infty} \frac{3x}{\sqrt{4x^2 + 3x} + 2x}
\]

### Step 4: Factor \( x \) from the Terms
In the fraction, factor \( x \) from the numerator and also from the terms in the denominator:
\[
= \lim_{x \to +\infty} \frac{3x}{x \left( \sqrt{4 + \frac{3}{x}} + 2 \right)}
\]
Simplifying by canceling \( x \) in the numerator and denominator:
\[
= \lim_{x \to +\infty} \frac{3}{\sqrt{4 + \frac{3}{x}} + 2}
\]

### Step 5: Evaluate the Limit as \( x \to +\infty \)
As \( x \to +\infty \), the term \( \frac{3}{x} \) approaches zero. Therefore:
\[
= \frac{3}{\sqrt{4 + 0} + 2} = \frac{3}{2 + 2} = \frac{3}{4}
\]

### Final Answer
\[
\lim_{x \to +\infty} \left( \sqrt{4x^2 + 3x} - 2x \right) = \frac{3}{4}
\]

Would you like additional clarification on any of the steps?

### Related Questions
1. What would the limit be if the term inside the square root were different, like \( 5x^2 + 3x \)?
2. How does this process change if we approach \( x \to -\infty \) instead?
3. How can we apply L'Hôpital's Rule to this limit problem?
4. What happens if the 3x term inside the square root is replaced with a higher power of x?
5. How would you approach similar limits with terms like \( \sqrt{x^2 + x} - x \)?

### Tip:
When dealing with limits at infinity involving square roots, rationalizing by the conjugate often simplifies complex expressions effectively.

Learn how to find the limit of lim x -> +inf, sqrt(4x^2 + 3x) - 2x using rationalization and algebraic manipulation. This solution covers key calculus concepts like limits at infinity and rationalizing square root expressions, ideal for high school students in Grades 11-12.

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