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Calcula el \( \lim _{x \rightarrow 2} \sqrt[3]{3 x^{2}-2 x+7} \)

Ask by Donnelly Robbins. in Mexico
Jan 23,2025

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Answer

El límite es \( \sqrt[3]{15} \), que es aproximadamente 2.466.

Solution

Para calcular el límite: \[ \lim _{x \rightarrow 2} \sqrt[3]{3 x^{2}-2 x+7} \] Vamos a evaluar la expresión dentro de la raíz cúbica en \( x = 2 \): 1. Sustituimos \( x \) por 2 en la expresión \( 3x^{2} - 2x + 7 \): \[ 3(2)^{2} - 2(2) + 7 = 3 \times 4 - 4 + 7 = 12 - 4 + 7 = 15 \] 2. Ahora, tomamos la raíz cúbica de 15: \[ \sqrt[3]{15} \] Por lo tanto, el límite es: \[ \lim _{x \rightarrow 2} \sqrt[3]{3 x^{2}-2 x+7} = \sqrt[3]{15} \] Si deseas una aproximación decimal, \(\sqrt[3]{15} \approx 2.466\).

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Para calcular el límite \( \lim _{x \rightarrow 2} \sqrt[3]{3 x^{2}-2 x+7} \), primero evaluamos la expresión dentro de la raíz cúbica en \( x = 2 \): \[ 3(2)^2 - 2(2) + 7 = 3(4) - 4 + 7 = 12 - 4 + 7 = 15. \] Ahora, sustituimos esto en el límite: \[ \lim _{x \rightarrow 2} \sqrt[3]{3 x^{2}-2 x+7} = \sqrt[3]{15}. \] Por lo tanto, el límite es \[ \sqrt[3]{15}. \]

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