Math Problem Statement

Can you solve without L'Hôpital?

Solution

Certainly! Let's solve the limit without using L'Hôpital's rule by focusing on the dominant terms and simplifying the expression directly.

The limit to solve is:

limx(2x2x333x223x25)\lim_{{x \to \infty}} \left( \frac{\sqrt[3]{2x^2 - x^3}}{3x^2 - 2\sqrt{3x^2} - 5} \right)

Step-by-Step Solution:

  1. Simplify the Numerator and Denominator by Factoring:

    We start by identifying the highest power of xx in both the numerator and the denominator.

    In the numerator, 2x2x32x^2 - x^3 can be written as: [ \sqrt[3]{2x^2 - x^3} = \sqrt[3]{-x^3 (1 - \frac{2x^2}{x^3})} = \sqrt[3]{-x^3 (1 - \frac{2}{x})} ]

    For large xx, 2x\frac{2}{x} approaches 0, so: [ \sqrt[3]{2x^2 - x^3} \approx \sqrt[3]{-x^3} = -x ]

    In the denominator, 3x223x253x^2 - 2\sqrt{3x^2} - 5 can be simplified by dividing each term by x2x^2: 3x223x25=3x223x53x^2 - 2\sqrt{3x^2} - 5 = 3x^2 - 2\sqrt{3}x - 5 The dominant term here is 3x23x^2.

  2. Rewrite the Expression:

    Given the dominant terms, we rewrite the original limit expression as: 2x2x333x223x25x3x2\frac{\sqrt[3]{2x^2 - x^3}}{3x^2 - 2\sqrt{3x^2} - 5} \approx \frac{-x}{3x^2}

  3. Simplify the Fraction:

    Simplify the fraction: x3x2=13x\frac{-x}{3x^2} = \frac{-1}{3x}

  4. Evaluate the Limit:

    As xx \to \infty: limx13x=0\lim_{{x \to \infty}} \frac{-1}{3x} = 0

Conclusion:

limx(2x2x333x223x25)=0\lim_{{x \to \infty}} \left( \frac{\sqrt[3]{2x^2 - x^3}}{3x^2 - 2\sqrt{3x^2} - 5} \right) = 0

Would you like any additional details or have any questions about this solution?

Here are 5 questions you might ask next:

  1. How can we determine the dominant terms in a function as xx \to \infty?
  2. What techniques can be used to simplify complex fractions?
  3. How do different powers of xx affect the behavior of a function at infinity?
  4. Can you explain how to handle limits involving different types of roots?
  5. What are the key differences between using algebraic simplifications and L'Hôpital's rule for limits?

Tip: When dealing with limits, especially as xx \to \infty, focusing on the highest power terms in both the numerator and denominator can often simplify the problem and lead you to the correct answer efficiently.

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Math Problem Analysis

Mathematical Concepts

Limits
Infinity
Simplification of Expressions

Formulas

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Theorems

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Suitable Grade Level

Undergraduate