Q103 / 160 · AP EAMCET 2006 · Mathematics

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

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

Solution

$\lim _{x \rightarrow \infty}\left[\sqrt{x^2+2 x-1}-x\right]$ $=\lim _{x \rightarrow \infty}\left[\frac{\left.\sqrt{\left(x^2+2 x-1\right.}-x\right)\left(\sqrt{x^2+2 x-1}+x\right)}{\sqrt{x^2+2 x-1}+x}\right]$ $=\lim _{x \rightarrow \infty}\left[\frac{x^2+2 x-1-x^2}{\sqrt{x^2+2 x-1}+x}\right]$ $=\lim _{x \rightarrow \infty}\left[\frac{2-\frac{1}{x}}{\sqrt{1+\frac{2}{x}-\frac{1}{x^2}+1}}\right]$ $=\frac{2}{2}=1$

Hence, the correct answer is option: 'D'.

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