The value of the limit lim _{x rightarrow-inf | vedclass

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(D) To evaluate the limit $L = \lim _{x \rightarrow-\infty}\left(\sqrt{4 x^2-x}+2 x\right)$,we rationalize the expression.Multiply and divide by the c

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$\frac{1}{4}$

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(D) To evaluate the limit $L = \lim _{x \rightarrow-\infty}\left(\sqrt{4 x^2-x}+2 x\right)$,we rationalize the expression.Multiply and divide by the conjugate $\left(\sqrt{4 x^2-x}-2 x\right)$:$L = \lim _{x \rightarrow-\infty} \frac{(\sqrt{4 x^2-x}+2 x)(\sqrt{4 x^2-x}-2 x)}{\sqrt{4 x^2-x}-2 x}$$L = \lim _{x \rightarrow-\infty} \frac{4 x^2 - x - 4x^2}{\sqrt{4 x^2-x}-2 x} = \lim _{x \rightarrow-\infty} \frac{-x}{\sqrt{x^2(4 - \frac{1}{x})} - 2x}$Since $x \rightarrow -\infty$,we have $\sqrt{x^2} = |x| = -x$:$L = \lim _{x \rightarrow-\infty} \frac{-x}{-x\sqrt{4 - \frac{1}{x}} - 2x} = \lim _{x \rightarrow-\infty} \frac{-x}{-x(\sqrt{4 - \frac{1}{x}} + 2)}$$L = \lim _{x \rightarrow-\infty} \frac{1}{\sqrt{4 - \frac{1}{x}} + 2} = \frac{1}{\sqrt{4-0} + 2} = \frac{1}{2+2} = \frac{1}{4}$.

The value of the limit $\lim _{x \rightarrow-\infty}\left(\sqrt{4 x^2-x}+2 x\right)$ is

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