\(\lim _{x\rightarrow \infty }(\sqrt{\frac{\left(8x^{2}-3\right)}{\left(2x^{2}+x\right)}})\)

We are tasked with finding the limit:

\[
\lim_{x \to \infty} \sqrt{\frac{8x^2 - 3}{2x^2 + x}}
\]

### Step 1: Simplify the expression inside the square root.

The expression inside the square root is:

\[
\frac{8x^2 - 3}{2x^2 + x}
\]

To handle limits as \(x \to \infty\), it helps to factor out the highest power of \(x\) from both the numerator and the denominator. In this case, \(x^2\) is the highest power.

\[
\frac{8x^2 - 3}{2x^2 + x} = \frac{x^2(8 - \frac{3}{x^2})}{x^2(2 + \frac{1}{x})}
\]

Now, we can cancel out \(x^2\) from both the numerator and the denominator:

\[
= \frac{8 - \frac{3}{x^2}}{2 + \frac{1}{x}}
\]

### Step 2: Take the limit as \(x \to \infty\).

As \(x \to \infty\), the terms \(\frac{3}{x^2}\) and \(\frac{1}{x}\) approach 0. Therefore, the expression simplifies to:

\[
\frac{8 - 0}{2 + 0} = \frac{8}{2} = 4
\]

### Step 3: Apply the square root.

Now, apply the square root to the result:

\[
\sqrt{4} = 2
\]

### Final Answer:

\[
\lim_{x \to \infty} \sqrt{\frac{8x^2 - 3}{2x^2 + x}} = 2
\]

Would you like further details or explanations on any of the steps?

Here are 5 related questions you might explore:
1. How would the limit change if the coefficient of \(x^2\) in the numerator were different?
2. What is the effect of the lower degree terms like \(x\) and constants on limits at infinity?
3. How do square roots interact with limits in general?
4. What if \(x \to -\infty\) instead of \(x \to \infty\) for this limit?
5. How do you handle limits of more complicated expressions involving polynomials?

**Tip:** When calculating limits at infinity, focus on the highest degree terms, as they dominate the behavior of the function.

lim _{x \rightarrow \infty }(\sqrt{\frac{\left(8x^{2}-3\right)}{\left(2x^{2}+x\right)}})

This page provides a step-by-step solution for finding the limit of the expression sqrt((8x^2 - 3) / (2x^2 + x)) as x approaches infinity. The problem involves simplifying polynomials and applying limits. Suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation