lim _{x rightarrow infty}left[sqrt{x^2+2 x-1} | vedclass

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(D) To evaluate the limit $\lim _{x \rightarrow \infty}\left[\sqrt{x^2+2 x-1}-x\right]$,we rationalize the expression by multiplying and dividing by t

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(D) To evaluate the limit $\lim _{x \rightarrow \infty}\left[\sqrt{x^2+2 x-1}-x\right]$,we rationalize the expression by multiplying and dividing by the conjugate $\left(\sqrt{x^2+2 x-1}+x\right)$:$\lim _{x}$ ${\rightarrow \infty}\left[\frac{\left(\sqrt{x^2+2 x-1}-x\right)\left(\sqrt{x^2+2 x-1}+x\right)}{\sqrt{x^2+2 x-1}+x}\right]$$= \lim _{x \rightarrow \infty}\left[\frac{x^2+2 x-1-x^2}{\sqrt{x^2+2 x-1}+x}\right]$$= \lim _{x \rightarrow \infty}\left[\frac{2x-1}{\sqrt{x^2+2 x-1}+x}\right]$Divide the numerator and denominator by $x$:$= \lim _{x \rightarrow \infty}\left[\frac{2-\frac{1}{x}}{\sqrt{1+\frac{2}{x}-\frac{1}{x^2}}+1}\right]$As $x \rightarrow \infty$,$\frac{1}{x} \rightarrow 0$ and $\frac{1}{x^2} \rightarrow 0$:$= \frac{2-0}{\sqrt{1+0-0}+1} = \frac{2}{1+1} = \frac{2}{2} = 1$

$\lim _{x \rightarrow \infty}\left[\sqrt{x^2+2 x-1}-x\right]$ is equal to :

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