Correct Answer - Option 1 : 1
Calculation:
\(\rm \mathop {\lim}\limits_{x\rightarrow \infty }\frac{\sqrt{x^{2}+2x-1}}{x}\\=\rm \mathop {\lim}\limits_{x\rightarrow \infty }\frac{\sqrt{x^{2}\left ( 1+\frac{2}{x}-\frac{1}{x^{2}} \right )}}{x}\)
\(=\rm \mathop {\lim}\limits_{x\rightarrow \infty }\frac{x\sqrt{\left ( 1+\frac{2}{x}-\frac{1}{x} \right )}}{x}\)
\(=\rm \mathop {\lim}\limits_{x\rightarrow \infty }\sqrt{1+\frac{2}{x}-\frac{1}{x^{2}}}\)
Applying the limits
\(=1\)
\(\therefore \) Option 1 is correct.
