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\lim _{x \rightarrow \infty}\left(x-\sqrt{x^2+x}\right)

Option: 1

\frac{-1}{4}


Option: 2

\frac{-1}{2}


Option: 3

\frac{-1}{3}


Option: 4

\frac{-3}{4}


Answers (1)

best_answer

By multiplying the numerator and denominator by the conjugate of the equation within the square root, you may make the expression simpler:

\begin{aligned} & \lim _{x \rightarrow \infty}\left(x-\sqrt{x^2+x}\right) \\ & =\lim _{x \rightarrow \infty}\left(x-\sqrt{x^2+x}\right) \cdot \frac{x+\sqrt{x^2+x}}{x+\sqrt{x^2+x}} \end{aligned}

use the identity  (a-b)(a+b)=a^2-b^2 to simplify the numerator:

\begin{aligned} &\lim _{x \rightarrow \infty}\left(x-\sqrt{x^2+x}\right) \cdot \frac{x+\sqrt{x^2+x}}{x+\sqrt{x^2+x}}\\ &\lim _{x \rightarrow \infty} \frac{x^2-\left(x^2+x\right)}{x+\sqrt{x^2+x}}=\lim _{x \rightarrow \infty} \frac{-1}{1+\sqrt{1+\frac{1}{x}}} \end{aligned}

\begin{aligned} & =\frac{-1}{1+\sqrt{1+\frac{1}{\infty}}} \\ & =\frac{-1}{1+\sqrt{1+0}}=\frac{-1}{1+1} \\ & =\frac{-1}{2} \end{aligned}

 

 

Posted by

jitender.kumar

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