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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

Vedclass · Accepted Answer

$\frac{1}{4}$

Vedclass · Answer

(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

Similar Questions

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{2 \sqrt{2}-(\cos x+\sin x)^3}{1-\sin 2 x}=$

The value of $\mathop {\lim }\limits_{x \to \infty } (\sqrt {{a^2}{x^2} + ax + 1} - \sqrt {{a^2}{x^2} + 1})$ is

The true statement for $\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{{\sqrt {2 + 3x} - \sqrt {2 - 3x} }}$ is

The value of $\lim _{x \rightarrow 1} \frac{x+x^2+\ldots+x^n-n}{x-1}$ is

The value of $\mathop {\lim }\limits_{x \to - 1} \frac{{{x^2} + 3x + 2}}{{{x^2} + 4x + 3}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module