A) \[\infty \]
B) \[\frac{1}{2}\]
C) 4
D) 1
Correct Answer: D
Solution :
\[\underset{x\to \infty }{\mathop{\lim }}\,[\sqrt{{{x}^{2}}+2x-1}-x]\] \[=\underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{(\sqrt{{{x}^{2}}+2x-1}-x)(\sqrt{{{x}^{2}}+2x-1}+x)}{\sqrt{{{x}^{2}}+2x-1}+x} \right]\] \[=\underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{{{x}^{2}}+2x-1-{{x}^{2}}}{\sqrt{{{x}^{2}}+2x-1}+x} \right]\] \[=\underset{x\to \infty }{\mathop{\lim }}\,\left[ \frac{2-\frac{1}{x}}{\sqrt{1+\frac{2}{x}-\frac{1}{{{x}^{2}}}+1}} \right]\] \[=\frac{2}{2}=1\]You need to login to perform this action.
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