Barnett Murphy
03/12/2024 · Middle School
b) \( \lim _{x \rightarrow+\infty}\left(e^{-5 x}\left(3 x^{2}-x\right)\right) \)
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Calculate the limit \( \lim _{x \rightarrow+\infty}\left(e^{-5 x}\left(3 x^{2}-x\right)\right) \).
Evaluate the limit by following steps:
- step0: Evaluate using L'Hopital's rule:
\(\lim _{x\rightarrow +\infty}\left(e^{-5x}\left(3x^{2}-x\right)\right)\)
- step1: Multiply the terms:
\(\lim _{x\rightarrow +\infty}\left(\left(3x^{2}-x\right)e^{-5x}\right)\)
- step2: Transform the expression:
\(\lim _{x\rightarrow +\infty}\left(\frac{3x^{2}-x}{e^{5x}}\right)\)
- step3: Use the L'Hopital's rule:
\(\lim _{x\rightarrow +\infty}\left(\frac{\frac{d}{dx}\left(3x^{2}-x\right)}{\frac{d}{dx}\left(e^{5x}\right)}\right)\)
- step4: Find the derivative:
\(\lim _{x\rightarrow +\infty}\left(\frac{6x-1}{5e^{5x}}\right)\)
- step5: Use the L'Hopital's rule:
\(\lim _{x\rightarrow +\infty}\left(\frac{\frac{d}{dx}\left(6x-1\right)}{\frac{d}{dx}\left(5e^{5x}\right)}\right)\)
- step6: Find the derivative:
\(\lim _{x\rightarrow +\infty}\left(\frac{6}{25e^{5x}}\right)\)
- step7: Rewrite the expression:
\(\frac{6}{\lim _{x\rightarrow +\infty}\left(25e^{5x}\right)}\)
- step8: Calculate:
\(\frac{6}{+\infty}\)
- step9: Divide the terms:
\(0\)
La expresión \( \lim _{x \rightarrow+\infty}\left(e^{-5 x}\left(3 x^{2}-x\right)\right) \) tiene un límite de 0.
Quick Answer
El límite es 0.
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