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Evaluate the limit \( \lim _{x \rightarrow \infty} \frac{11 x^{2}-2 x+6}{4 x+8} \)

Ask by Burton Schwartz. in the United States
Jan 29,2025

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Answer

The limit is \( +\infty \).

Solution

To evaluate the limit \[ \lim_{x \rightarrow \infty} \frac{11x^{2} - 2x + 6}{4x + 8}, \] let's analyze the behavior of the numerator and the denominator as \( x \) approaches infinity. 1. **Degree Comparison:** - **Numerator:** The highest power of \( x \) is \( x^2 \), so the degree is 2. - **Denominator:** The highest power of \( x \) is \( x \), so the degree is 1. 2. **Dominant Terms:** - As \( x \) becomes very large, the lower-degree terms (\( -2x + 6 \) in the numerator and \( 8 \) in the denominator) become negligible compared to the highest-degree terms. - Therefore, the expression behaves like: \[ \frac{11x^{2}}{4x} = \frac{11}{4}x \] 3. **Limit Evaluation:** - The simplified expression \( \frac{11}{4}x \) clearly tends to \( \infty \) as \( x \) approaches \( \infty \). Therefore, the original limit evaluates to positive infinity. **Answer:** The limit is \( +\infty \).

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The Deep Dive

To evaluate the limit \[ \lim _{x \rightarrow \infty} \frac{11 x^{2}-2 x+6}{4 x+8}, \] we can first identify the highest power of \(x\) in both the numerator and the denominator. The numerator has a highest power of \(x^2\) and the denominator a highest power of \(x^1\). To simplify, we can divide every term in the numerator and denominator by \(x^2\): \[ \lim _{x \rightarrow \infty} \frac{11 - \frac{2}{x} + \frac{6}{x^2}}{\frac{4}{x} + \frac{8}{x^2}}. \] As \(x\) approaches infinity, the terms \(\frac{2}{x}\), \(\frac{6}{x^2}\), \(\frac{4}{x}\), and \(\frac{8}{x^2}\) will approach zero. Thus, we can simplify the limit to: \[ \lim _{x \rightarrow \infty} \frac{11 - 0 + 0}{0 + 0} = \frac{11}{0}. \] Since the degree of the numerator is higher than that of the denominator, the limit approaches infinity. Therefore, we conclude that: \[ \lim _{x \rightarrow \infty} \frac{11 x^{2}-2 x+6}{4 x+8} = \infty. \]

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