I see that you've uploaded an image containing a mathematical expression. Let me extract the equation and solve it.

The given expression appears to be:

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

Let me solve this step by step.

Step 1: Simplify the numerator

The numerator is:

$\frac{2x^4 - 1 + 3x^2}{x}.$

Divide each term by $x$:

$\frac{2x^4}{x} - \frac{1}{x} + \frac{3x^2}{x} = 2x^3 - \frac{1}{x} + 3x.$

Thus, the numerator becomes $2x^3 + 3x - \frac{1}{x}$.

Step 2: Write the denominator

The denominator is:

$x\sqrt{x} - 4.$

Step 3: Combine the numerator and denominator

The overall expression becomes:

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

Step 4: Analyze dominant terms as $x \to \infty$

• In the numerator, the dominant term is $2x^3$ since $x^3$ grows faster than $x$ or $-\frac{1}{x}$.
• In the denominator, the dominant term is $x\sqrt{x} = x^{3/2}$, as it grows much faster than $-4$.

So, the expression simplifies to:

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

Step 5: Simplify the dominant terms

Simplify $\frac{2x^3}{x\sqrt{x}}$:

$\frac{2x^3}{x\sqrt{x}} = \frac{2x^3}{x^{3/2}} = 2x^{3 - 3/2} = 2x^{3/2}.$

Step 6: Evaluate the limit

As $x \to \infty$, $2x^{3/2} \to \infty$. Therefore, the limit diverges to infinity:

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

Final Answer:

$\boxed{\infty}$

Do you want a breakdown of any step or further clarifications?

Related Questions:

1. How do we determine the dominant term in a rational expression as $x \to \infty$?
2. Why is $x^{3/2}$ the dominant term in the denominator?
3. Can limits like these approach finite values instead of diverging?
4. How does L'Hopital's Rule apply in cases of infinity/infinity indeterminate forms?
5. How does dividing by the highest power of $x$ simplify such limits?

Tip:

Always identify and focus on the dominant terms for large values of $x$ to simplify expressions and evaluate limits efficiently.