Multiple Choice 1 point Evaluate \( \lim _{x \rightarrow \infty}(\sqrt{x+1}-\sqrt{x}) \) -1 0 \( \infty \) 1
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Bonus Knowledge
To evaluate the limit \( \lim _{x \rightarrow \infty}(\sqrt{x+1}-\sqrt{x}) \), we can manipulate the expression to simplify it. We can multiply and divide by the conjugate: \[ \lim _{x \rightarrow \infty} \left(\sqrt{x+1} - \sqrt{x}\right) \cdot \frac{\sqrt{x+1} + \sqrt{x}}{\sqrt{x+1} + \sqrt{x}} = \lim _{x \rightarrow \infty} \frac{(x + 1) - x}{\sqrt{x+1} + \sqrt{x}} = \lim _{x \rightarrow \infty} \frac{1}{\sqrt{x+1} + \sqrt{x}} \] As \( x \) approaches infinity, \( \sqrt{x+1} \) and \( \sqrt{x} \) both approach \( \sqrt{x} \), so we have: \[ \lim_{x \rightarrow \infty} \frac{1}{\sqrt{x+1} + \sqrt{x}} = \lim_{x \rightarrow \infty} \frac{1}{\sqrt{x} + \sqrt{x}} = \lim_{x \rightarrow \infty} \frac{1}{2\sqrt{x}} = 0 \] Thus, the limit is \( 0 \). Therefore, the correct answer is \( 0 \).
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