Question:
\( \lim_{x\to\infty} \left(\sqrt{x^2+1} - \sqrt{x^2-1}\right) \)
\( \lim_{x\to\infty} \left(\sqrt{x^2+1} - \sqrt{x^2-1}\right) \)
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Use conjugate for expressions with square roots.
Updated On: May 1, 2026
- \( -1 \)
- \( 1 \)
- \( 0 \)
- \( 2 \)
- \( 4 \)
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The Correct Option is B
Solution and Explanation
Concept: Rationalization.
Step 1: Multiply conjugate.
\[ \frac{(x^2+1)-(x^2-1)}{\sqrt{x^2+1}+\sqrt{x^2-1}} \]
Step 2: Simplify numerator.
\[ = \frac{2}{\sqrt{x^2+1}+\sqrt{x^2-1}} \]
Step 3: Divide by \( x \).
Step 4: Limit: \[ = \frac{2}{x(1+1)} = \frac{1}{x} \]
Step 5: As \( x\to\infty \): \[ =0 \]
Step 1: Multiply conjugate.
\[ \frac{(x^2+1)-(x^2-1)}{\sqrt{x^2+1}+\sqrt{x^2-1}} \]
Step 2: Simplify numerator.
\[ = \frac{2}{\sqrt{x^2+1}+\sqrt{x^2-1}} \]
Step 3: Divide by \( x \).
Step 4: Limit: \[ = \frac{2}{x(1+1)} = \frac{1}{x} \]
Step 5: As \( x\to\infty \): \[ =0 \]
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