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Determine \( \lim _{x \rightarrow-\infty} \frac{3 x^{2}+12}{x^{2}+x-12} \)

Ask by Young Nunez. in the United States
Mar 14,2025

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The limit is 3.

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To evaluate the limit \[ \lim_{x \rightarrow -\infty} \frac{3x^{2} + 12}{x^{2} + x - 12}, \] we start by dividing the numerator and the denominator by \(x^2\), the highest power of \(x\) in both the numerator and the denominator: \[ \frac{3x^{2} + 12}{x^{2} + x - 12} = \frac{3 + \frac{12}{x^{2}}}{1 + \frac{1}{x} - \frac{12}{x^{2}}}. \] Now, as \(x \to -\infty\), the terms \(\frac{12}{x^{2}}\) and \(\frac{1}{x}\) approach \(0\): - \(\frac{12}{x^{2}} \to 0\), - \(\frac{1}{x} \to 0\). Thus, the limit simplifies to: \[ \lim_{x \rightarrow -\infty} \frac{3 + 0}{1 + 0 - 0} = \frac{3}{1} = 3. \] Therefore, the final result is: \[ \boxed{3}. \]

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