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Question
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\] is equal to
Options
\[\frac{1}{2}\]
2
0
1
Solution
\[\frac{1}{2}\]
\[\lim_{x \to 0} \frac{\sqrt{1 + x} - 1}{x}\]
\[ = \lim_{x \to 0} \frac{\left( \sqrt{1 + x} - 1 \right) \left( \sqrt{1 + x} + 1 \right)}{\left( \sqrt{1 + x} + 1 \right) x}\]
\[ = \lim_{x \to 0} \frac{1 + x - 1}{x\sqrt{1 + x + 1}}\]
\[ = \lim_{x \to 0} \frac{1}{x\left( \sqrt{1 + x} + 1 \right)}\]
\[ = \frac{1}{2}\]
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