The image you provided is a mathematical problem involving limits. The problem asks to find the limit:

$\lim_{x \to \infty} \frac{x^2 + 3x}{\sqrt{2x^4 + 3 - 4x}}$

The solution proceeds step by step. Here's a breakdown:

Step 1: Expressing the equation

We are given the limit expression: $\frac{x^2 + 3x}{\sqrt{2x^4 + 3 - 4x}}$ The goal is to simplify this expression for $x \to \infty$.

Step 2: Factoring out dominant terms

• In the numerator $x^2 + 3x$, the term $x^2$ dominates as $x \to \infty$.
• In the denominator, the dominant term inside the square root is $2x^4$ because as $x$ grows large, other terms become negligible.

We factor out $x^2$ from both the numerator and denominator to make the expression easier to evaluate as $x \to \infty$:

\frac{x^2 + 3x}{\sqrt{2x^4 + 3 - 4x}} = \frac{x^2(1 + \frac{3}{x})}{x^2\sqrt{2 + \frac{3}{x^4} - \frac{4}{x^3}}}}

Step 3: Simplification

As $x \to \infty$, terms involving $\frac{1}{x}, \frac{1}{x^3},$ and $\frac{1}{x^4}$ tend to zero. Therefore, the expression simplifies to:

$\frac{1}{\sqrt{2}}$

Step 4: Conclusion

Thus, the limit is:

$\lim_{x \to \infty} \frac{x^2 + 3x}{\sqrt{2x^4 + 3 - 4x}} = \frac{1}{\sqrt{2}}$

Would you like more details on any specific step or the rules applied here?

Further Questions:

1. Why do we focus on the dominant terms as $x \to \infty$?
2. What happens to smaller order terms as $x$ approaches infinity?
3. How does the square root affect the terms in the denominator?
4. Could this limit be evaluated using L'Hopital's rule instead?
5. What does the result $\frac{1}{\sqrt{2}}$ imply about the growth rates of the numerator and denominator?

Tip:

When evaluating limits as $x \to \infty$, always identify and simplify dominant terms, as smaller ones become negligible!