TS EAMCET · Maths · Limits
\(\lim _{x \rightarrow \infty}\left[\sqrt{x^2+2 x-1}-x\right]\) is equal to :
- A \(\infty\)
- B \(\frac{1}{2}\)
- C 4
- D 1
Answer & Solution
Correct Answer
(D) 1
Step-by-step Solution
Detailed explanation
\(\lim _{x \rightarrow \infty}\left[\sqrt{x^2+2 x-1}-x\right]\) \(=\lim _{x \rightarrow \infty}\left[\frac{\left.\sqrt{\left(x^2+2 x-1\right.}-x\right)\left(\sqrt{x^2+2 x-1}+x\right)}{\sqrt{x^2+2 x-1}+x}\right]\)…
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