lim _{x rightarrow infty}left(sqrt[3]{x^3+4 x | vedclass

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(A) We want to evaluate $L = \lim _{x \rightarrow \infty}\left(\sqrt[3]{x^3+4 x^2}-\sqrt{x^2-3 x}\right)$.First,factor out $x$ from each term:$L = \li

Vedclass · Accepted Answer

$\frac{17}{6}$

Vedclass · Answer

(A) We want to evaluate $L = \lim _{x \rightarrow \infty}\left(\sqrt[3]{x^3+4 x^2}-\sqrt{x^2-3 x}\right)$.First,factor out $x$ from each term:$L = \lim _{x \rightarrow \infty} \left( x(1+\frac{4}{x})^{1/3} - x(1-\frac{3}{x})^{1/2} \right)$.Using the binomial expansion $(1+u)^n \approx 1 + nu + \frac{n(n-1)}{2}u^2$ for small $u$:$(1+\frac{4}{x})^{1/3} \approx 1 + \frac{1}{3}(\frac{4}{x}) = 1 + \frac{4}{3x}$.$(1-\frac{3}{x})^{1/2} \approx 1 + \frac{1}{2}(-\frac{3}{x}) = 1 - \frac{3}{2x}$.Substituting these into the limit:$L = \lim _{x \rightarrow \infty} \left( x(1 + \frac{4}{3x}) - x(1 - \frac{3}{2x}) \right)$.$L = \lim _{x \rightarrow \infty} \left( x + \frac{4}{3} - x + \frac{3}{2} \right)$.$L = \frac{4}{3} + \frac{3}{2} = \frac{8+9}{6} = \frac{17}{6}$.

$\lim _{x \rightarrow \infty}\left(\sqrt[3]{x^3+4 x^2}-\sqrt{x^2-3 x}\right)=$

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