Question:
\( \lim_{x \to 0} \frac{\sqrt{1+2x} - 1}{x} \) is
\( \lim_{x \to 0} \frac{\sqrt{1+2x} - 1}{x} \) is
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Rationalization is very useful for limits involving roots.
Updated On: May 1, 2026
- \( 0 \)
- \( -1 \)
- \( \frac{1}{2} \)
- \( 1 \)
- \( -\frac{1}{2} \)
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The Correct Option is D
Solution and Explanation
Concept:
Use rationalization or derivative definition.
Step 1: Multiply numerator and denominator by conjugate.
\[ \frac{\sqrt{1+2x}-1}{x} \cdot \frac{\sqrt{1+2x}+1}{\sqrt{1+2x}+1} \]
Step 2: Simplify numerator.
\[ = \frac{(1+2x) - 1}{x(\sqrt{1+2x}+1)} = \frac{2x}{x(\sqrt{1+2x}+1)} \]
Step 3: Cancel \( x \).
\[ = \frac{2}{\sqrt{1+2x}+1} \]
Step 4: Take limit as \( x \to 0 \).
\[ = \frac{2}{1+1} = 1 \]
Step 5: Final answer.
\[ 1 \]
Step 1: Multiply numerator and denominator by conjugate.
\[ \frac{\sqrt{1+2x}-1}{x} \cdot \frac{\sqrt{1+2x}+1}{\sqrt{1+2x}+1} \]
Step 2: Simplify numerator.
\[ = \frac{(1+2x) - 1}{x(\sqrt{1+2x}+1)} = \frac{2x}{x(\sqrt{1+2x}+1)} \]
Step 3: Cancel \( x \).
\[ = \frac{2}{\sqrt{1+2x}+1} \]
Step 4: Take limit as \( x \to 0 \).
\[ = \frac{2}{1+1} = 1 \]
Step 5: Final answer.
\[ 1 \]
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