Question

$limi{t}_{x\to \infty }\left(\sqrt{x^2+2x-1}\right)-x=$

• A. $\infty$
• B. $\frac{1}{2}$
• C. $4$
• D. $1$

Answer 1

The correct option is D

$1$

Explanation for the correct option:

Evaluating the ${\text{limit}}_{x\to \infty }\left(\sqrt{x^2+2x-1}\right)-x$

when we substitute infinite, we get the value $\left(-\infty ,\infty \right)$

$$\begin{gathered} =limi{t}_{x\to \infty }\left(\sqrt{x^2+2x-1}\right)-x \\ =limi{t}_{x\to \infty }\left[\left(\left(\sqrt{x^2+2x-1}\right)-x\right)\times \frac{\left(\sqrt{x^2+2x-1}\right)+x}{\left(\sqrt{x^2+2x-1}\right)+x}\right]\left[\mathbf{\text{multiply and divide}}\mathit{b}\mathit{y}\mathbf{}\mathbf{\left(}\sqrt{{\mathbf{x}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{x}\mathbf{-}\mathbf{1}}\mathbf{\right)}+x\right] \\ =limi{t}_{x\to \infty }\left[\frac{{\left(\sqrt{x^2+2x-1}\right)}^2-x^2}{\left(\sqrt{x^2+2x-1}\right)+x}\right] \\ =limi{t}_{x\to \infty }\left[\frac{x^2+2x-1-x^2}{\left(\sqrt{x^2+2x-1}\right)+x}\right] \\ =limi{t}_{x\to \infty }\left[\frac{x\left(2-{\displaystyle \frac{1}{x}}\right)}{\left(\sqrt{x^2\left(1+{\displaystyle \frac{2}{x}}-{\displaystyle \frac{1}{x^2}}\right)}\right)+x}\right] \\ =limi{t}_{x\to \infty }\left[\frac{x\left(2-{\displaystyle \frac{1}{x}}\right)}{\left(x\sqrt{\left(1+{\displaystyle \frac{2}{x}}-{\displaystyle \frac{1}{x^2}}\right)}\right)+1}\right] \\ =\frac{\left(2-{\displaystyle \frac{1}{\infty }}\right)}{\left(\sqrt{\left(1+{\displaystyle \frac{2}{\infty }}-{\displaystyle \frac{1}{{\infty }^2}}\right)}\right)+1} \\ =\frac{\left(2-{\displaystyle 0}\right)}{\left(\sqrt{\left(1+{\displaystyle 0}-{\displaystyle 0}\right)}\right)+1}\left[\because \frac{1}{\infty }=0\right] \\ =\frac{2}{2} \\ =1 \end{gathered}$$

Hence, option (D) is the correct answer