Search any question & find its solution
Question:
Answered & Verified by Expert
$\lim _{x \rightarrow \infty}\left[\sqrt{x^2+2 x-1}-x\right]$ is equal to :
Options:
Solution:
1958 Upvotes
Verified Answer
The correct answer is:
1
$\lim _{x \rightarrow \infty}\left[\sqrt{x^2+2 x-1}-x\right]$
$=\lim _{x \rightarrow \infty}\left[\frac{\left.\sqrt{\left(x^2+2 x-1\right.}-x\right)\left(\sqrt{x^2+2 x-1}+x\right)}{\sqrt{x^2+2 x-1}+x}\right]$
$=\lim _{x \rightarrow \infty}\left[\frac{x^2+2 x-1-x^2}{\sqrt{x^2+2 x-1}+x}\right]$
$=\lim _{x \rightarrow \infty}\left[\frac{2-\frac{1}{x}}{\sqrt{1+\frac{2}{x}-\frac{1}{x^2}+1}}\right]$
$=\frac{2}{2}=1$
$=\lim _{x \rightarrow \infty}\left[\frac{\left.\sqrt{\left(x^2+2 x-1\right.}-x\right)\left(\sqrt{x^2+2 x-1}+x\right)}{\sqrt{x^2+2 x-1}+x}\right]$
$=\lim _{x \rightarrow \infty}\left[\frac{x^2+2 x-1-x^2}{\sqrt{x^2+2 x-1}+x}\right]$
$=\lim _{x \rightarrow \infty}\left[\frac{2-\frac{1}{x}}{\sqrt{1+\frac{2}{x}-\frac{1}{x^2}+1}}\right]$
$=\frac{2}{2}=1$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.