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Question
\[\lim_{x \to \infty} \frac{\sqrt{x^2 - 1}}{2x + 1}\]
Options
1
0
−1
1/2
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Solution
1/2
\[ = \lim_{x \to \infty} \frac{\sqrt{1 - \frac{1}{x^2}}}{2 + \frac{1}{x}}\]
\[\text{ Dividing the numerator and the denominator by x, we get }: \]
\[ = \frac{\sqrt{1 - 0}}{2 + 0}\]
\[ = \frac{1}{2}\]
Hence, the correct answer is d.
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