The given problem is:

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

Solution:

Step 1: Simplify the square root expression

In the numerator, factor out $x^4$ from the terms inside the square root:

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

Taking $x^2$ outside the square root:

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

Step 2: Rewrite the fraction

The fraction becomes:

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

Step 3: Factor $x$ out of the denominator

In the denominator $-2x + 1$, factor out $x$:

$-2x + 1 = x(-2 + \frac{1}{x}).$

Thus, the fraction simplifies to:

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

Cancel one $x$ term from numerator and denominator:

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

Step 4: Evaluate the limit as $x \to \infty$

• As $x \to \infty$, the terms $\frac{1}{x}$ and $\frac{1}{x^2}$ approach $0$.
• The numerator simplifies to $x \cdot \sqrt{1 + 0 - 0} = x \cdot 1 = x$.
• The denominator simplifies to $-2 + 0 = -2$.

Thus, the fraction becomes:

$\frac{x}{-2} = -\frac{x}{2}.$

Conclusion:

As $x \to \infty$, the value of the limit is:

$\boxed{-\infty}.$

Let me know if you need further clarifications!

Related Questions:

1. How do you simplify square roots when factoring out powers of $x$?
2. What happens to terms involving $\frac{1}{x}$ and $\frac{1}{x^2}$ as $x \to \infty$?
3. Why is factoring the largest power of $x$ useful in limits at infinity?
4. What is the effect of a negative sign in the denominator on the result?
5. How does L'Hopital's Rule apply to similar limits?

Tip:

When evaluating limits at infinity, always simplify by factoring out the highest power of $x$ to avoid dealing with undefined forms.