The given limit is:

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

To evaluate this limit, let's break it down step by step:

Step 1: Simplify the denominator

The denominator is $\frac{3}{x^{\frac{1}{2}}} = 3x^{-\frac{1}{2}}$. So, the expression becomes:

$\frac{2x\sqrt{x} - x + 3}{3x^{-\frac{1}{2}}}$

Step 2: Multiply numerator and denominator by $x^{\frac{1}{2}}$

We multiply both the numerator and the denominator by $x^{\frac{1}{2}}$ to eliminate the negative exponent in the denominator:

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

Simplifying the terms in the numerator:

• $x^{\frac{1}{2}} \cdot 2x\sqrt{x} = 2x^{\frac{3}{2}} \cdot x^{\frac{1}{2}} = 2x^2$
• $x^{\frac{1}{2}} \cdot (-x) = -x^{\frac{3}{2}}$
• $x^{\frac{1}{2}} \cdot 3 = 3x^{\frac{1}{2}}$

So the expression becomes:

$\frac{2x^2 - x^{\frac{3}{2}} + 3x^{\frac{1}{2}}}{3}$

Step 3: Analyze the limit as $x \to \infty$

As $x$ approaches infinity, we observe the terms in the numerator:

• The term $2x^2$ dominates because it has the highest power of $x$.
• The other terms ($-x^{\frac{3}{2}}$ and $3x^{\frac{1}{2}}$) become negligible as $x \to \infty$.

Thus, the expression approaches:

$\lim_{x \to \infty} \frac{2x^2}{3} = \infty$

Conclusion:

The limit is infinite.

$\boxed{\infty}$

Would you like further clarification on any step, or to explore related problems?

Here are 5 related questions:

1. What happens if the power of $x$ in the denominator is higher than in the numerator?
2. How do limits behave when approaching $0$ instead of infinity?
3. Can L'Hopital's rule be used here? Why or why not?
4. How does multiplying by $x^{\frac{1}{2}}$ help simplify the problem?
5. What techniques can be applied to solve indeterminate limits?

Tip: When evaluating limits at infinity, always focus on the highest power terms in both the numerator and denominator to simplify the expression.