Question:
The value of \(\lim_{x \to 5} \left( \frac{25 - x^2}{4 - \sqrt{x^2 - 9}} \right)\) is:
The value of \(\lim_{x \to 5} \left( \frac{25 - x^2}{4 - \sqrt{x^2 - 9}} \right)\) is:
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Rationalization helps eliminate indeterminate forms quickly.
Updated On: Apr 30, 2026
- \(32 \)
- \(16 \)
- \(8 \)
- \(4 \)
- \(0 \)
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Verified By Collegedunia
The Correct Option is C
Solution and Explanation
Concept:
Use rationalization when square roots are involved.
Step 1: Factor numerator.
\[ 25 - x^2 = (5-x)(5+x) \]
Step 2: Multiply by conjugate.
\[ \frac{(25-x^2)(4+\sqrt{x^2-9})}{(4-\sqrt{x^2-9})(4+\sqrt{x^2-9})} \] \[ = \frac{(25-x^2)(4+\sqrt{x^2-9})}{16 - (x^2-9)} = \frac{(25-x^2)(4+\sqrt{x^2-9})}{25 - x^2} \]
Step 3: Cancel terms.
\[ = 4 + \sqrt{x^2-9} \]
Step 4: Substitute limit.
\[ = 4 + \sqrt{25 - 9} = 4 + 4 = 8 \]
Step 1: Factor numerator.
\[ 25 - x^2 = (5-x)(5+x) \]
Step 2: Multiply by conjugate.
\[ \frac{(25-x^2)(4+\sqrt{x^2-9})}{(4-\sqrt{x^2-9})(4+\sqrt{x^2-9})} \] \[ = \frac{(25-x^2)(4+\sqrt{x^2-9})}{16 - (x^2-9)} = \frac{(25-x^2)(4+\sqrt{x^2-9})}{25 - x^2} \]
Step 3: Cancel terms.
\[ = 4 + \sqrt{x^2-9} \]
Step 4: Substitute limit.
\[ = 4 + \sqrt{25 - 9} = 4 + 4 = 8 \]
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