Limits at Infinity, Part I — Question 1

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Question 1

Evaluate the limit: limx→∞6x3−4x2+103x3+5x−7\lim_{x \to \infty} \frac{6x^3 - 4x^2 + 10}{3x^3 + 5x - 7}

Original worksheet page 1: question and worked solution for 2-7-001
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Question 1 - Solution

We are given: limx→∞6x3−4x2+103x3+5x−7\lim_{x \to \infty} \frac{6x^3 - 4x^2 + 10}{3x^3 + 5x - 7}

Step 1: Identify the dominant terms.

As x→∞x \to \infty, the highest degree terms dominate: Numerator: 6x36x^3 , Denominator: 3x33x^3

So we focus on: 6x33x3=2\frac{6x^3}{3x^3} = 2

Step 2: Divide all terms by x3x^3: limx→∞6−4x+10x33+5x2−7x3=63=2\lim_{x \to \infty} \frac{6 - \frac{4}{x} + \frac{10}{x^3}}{3 + \frac{5}{x^2} - \frac{7}{x^3}} = \frac{6}{3} = 2

Conclusion: limx→∞6x3−4x2+103x3+5x−7=2\boxed{\lim_{x \to \infty} \frac{6x^3 - 4x^2 + 10}{3x^3 + 5x - 7} = 2}

Original worksheet page 2: question and worked solution for 2-7-001

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