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The problem presented is the limit:

\[
\lim_{x \to \infty} \left( \sqrt{4x^2 + 2x} - \sqrt{4x^2 - 2x} \right)
\]

### Solution:

First, observe that both square roots have terms involving \(x^2\), which grow faster than the linear terms involving \(x\). As \(x \to \infty\), the expression becomes difficult to evaluate directly, so we apply an algebraic trick called **multiplying by the conjugate** to simplify the limit.

We multiply both the numerator and denominator by the conjugate of the expression:

\[
\left( \sqrt{4x^2 + 2x} - \sqrt{4x^2 - 2x} \right) \times \frac{\sqrt{4x^2 + 2x} + \sqrt{4x^2 - 2x}}{\sqrt{4x^2 + 2x} + \sqrt{4x^2 - 2x}}
\]

This simplifies the numerator using the difference of squares formula:

\[
= \frac{(4x^2 + 2x) - (4x^2 - 2x)}{\sqrt{4x^2 + 2x} + \sqrt{4x^2 - 2x}}
\]

Simplifying the numerator:

\[
(4x^2 + 2x) - (4x^2 - 2x) = 4x
\]

So the expression becomes:

\[
\frac{4x}{\sqrt{4x^2 + 2x} + \sqrt{4x^2 - 2x}}
\]

### Now, let's analyze the denominator as \(x \to \infty\):

In both square roots, the \(4x^2\) term dominates, so we factor \(x^2\) out of each square root:

\[
\sqrt{4x^2 + 2x} = x\sqrt{4 + \frac{2}{x}}, \quad \sqrt{4x^2 - 2x} = x\sqrt{4 - \frac{2}{x}}
\]

As \(x \to \infty\), the terms involving \(\frac{2}{x}\) become negligible, so:

\[
\sqrt{4 + \frac{2}{x}} \to 2, \quad \sqrt{4 - \frac{2}{x}} \to 2
\]

Thus, the denominator becomes approximately:

\[
x(2 + 2) = 4x
\]

### Simplifying the entire expression:

\[
\frac{4x}{4x} = 1
\]

Therefore, the limit is:

\[
\lim_{x \to \infty} \left( \sqrt{4x^2 + 2x} - \sqrt{4x^2 - 2x} \right) = 1
\]

### Final Answer:
\[
\boxed{1}
\]

Let me know if you need further details or have any questions!

---

Here are some related questions:
1. What is the importance of the conjugate in simplifying limits involving square roots?
2. How does the growth rate of polynomial terms affect limits at infinity?
3. Can we apply L'Hopital's Rule to this limit, and would it give the same result?
4. How would the result change if the limit was taken as \(x \to -\infty\)?
5. What other techniques are commonly used to solve limits involving square roots?

**Tip**: When simplifying limits involving square roots at infinity, always check if multiplying by the conjugate can help simplify the expression!

lim_{x \to \infty} \left( \sqrt{4x^2 + 2x} - \sqrt{4x^2 - 2x} \right)

Learn how to solve the limit lim_{x \to \infty} (sqrt{4x^2 + 2x} - sqrt{4x^2 - 2x}) by using the conjugate method and simplifying the expression for large x. This problem focuses on limits involving square roots and applying the difference of squares formula.

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