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ㄹ. \( \lim _{x \rightarrow-\infty} \frac{\sqrt{9 x^{6}-x}}{x^{3}+1} \)

Ask by Ellis Clarke. in Mexico
Feb 04,2025

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Answer

El límite es -3.

Solution

Calculate the limit \( \lim_{x\rightarrow -\infty} \frac{\sqrt{9x^6-x}}{x^3+1} \). Evaluate the limit by following steps: - step0: Evaluate using transformations: \(\lim _{x\rightarrow -\infty}\left(\frac{\sqrt{9x^{6}-x}}{x^{3}+1}\right)\) - step1: Rewrite the expression: \(\lim _{x\rightarrow -\infty}\left(\frac{-\sqrt{9-\frac{1}{x^{5}}}\times x^{3}}{\left(1+\frac{1}{x^{3}}\right)x^{3}}\right)\) - step2: Simplify: \(\lim _{x\rightarrow -\infty}\left(\frac{-\sqrt{9-\frac{1}{x^{5}}}}{1+\frac{1}{x^{3}}}\right)\) - step3: Rewrite the expression: \(\frac{\lim _{x\rightarrow -\infty}\left(-\sqrt{9-\frac{1}{x^{5}}}\right)}{\lim _{x\rightarrow -\infty}\left(1+\frac{1}{x^{3}}\right)}\) - step4: Calculate: \(\frac{-3}{\lim _{x\rightarrow -\infty}\left(1+\frac{1}{x^{3}}\right)}\) - step5: Calculate: \(\frac{-3}{1}\) - step6: Divide the terms: \(-3\) La limitante de \( \frac{\sqrt{9x^6-x}}{x^3+1} \) cuando \( x \) se acerca a -∞ es -3.

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To solve the limit \( \lim _{x \rightarrow-\infty} \frac{\sqrt{9 x^{6}-x}}{x^{3}+1} \), we start by analyzing the expression as \( x \) approaches negative infinity. First, simplify the expression inside the square root and the denominator: \[ \sqrt{9 x^6 - x} \text{ can be approximated by } \sqrt{9 x^6} = 3|x^3| = 3(-x^3) = -3x^3 \text{ for large negative } x. \] The denominator behaves as follows: \[ x^3 + 1 \text{ approaches } x^3 \text{ for large negative } x. \] Now, substituting these simplifications into the limit, we have: \[ \lim_{x \rightarrow -\infty} \frac{\sqrt{9 x^6 - x}}{x^3 + 1} \approx \lim_{x \rightarrow -\infty} \frac{-3 x^3}{x^3} = \lim_{x \rightarrow -\infty} -3. \] Thus, the limit evaluates to: \[ \boxed{-3}. \]

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