To find the limit of the given expression

$\lim_{x \to -\infty} \left( \frac{1 - x^3}{x^2 + 5x} \right)^7$

Step 1: Divide numerator and denominator by the highest power of $x$ in the denominator.

The highest power in the denominator is $x^2$. So, we divide both the numerator and the denominator by $x^2$:

$\lim_{x \to -\infty} \left( \frac{\frac{1 - x^3}{x^2}}{\frac{x^2 + 5x}{x^2}} \right)^7$

Simplify each part:

• Numerator:

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

• Denominator:

$\frac{x^2 + 5x}{x^2} = 1 + \frac{5}{x}$

Now the limit becomes:

$\lim_{x \to -\infty} \left( \frac{\frac{1}{x^2} - x}{1 + \frac{5}{x}} \right)^7$

Step 2: Analyze the limit as $x$ approaches $-\infty$.

As $x$ goes to negative infinity:

• $\frac{1}{x^2} \to 0$
• $x \to -\infty$
• $\frac{5}{x} \to 0$

So, the expression simplifies to:

$\lim_{x \to -\infty} \left( \frac{-\infty}{1 + 0} \right)^7 = \left( -\infty \right)^7$

Since raising a negative number to an odd power preserves the negative sign:

$\left(-\infty\right)^7 = -\infty$

Final Answer:

$\lim_{x \to -\infty} \left( \frac{1 - x^3}{x^2 + 5x} \right)^7 = -\infty$

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Would you like me to go over any part of this step-by-step explanation?

Here are 5 related questions to consider:

1. What happens if the exponent were even instead of odd?
2. How would the limit change if $x$ approached $+\infty$?
3. Can you apply the same method to rational functions with higher powers?
4. How does the behavior of odd and even powers differ as $x$ goes to infinity?
5. How does dividing by the highest power of $x$ simplify limit calculations?

Tip: Always identify the dominant term when dealing with limits involving polynomials—it often determines the behavior of the entire expression!