Question:
\(\lim\limits_{t\rightarrow0}\frac{x}{\sqrt{9-x}-3}\) is equal to
\(\lim\limits_{t\rightarrow0}\frac{x}{\sqrt{9-x}-3}\) is equal to
Updated On: Apr 4, 2025
- 6
- 3
- -3
- -6
- 0
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The Correct Option is D
Solution and Explanation
We are given the limit:
\[ \lim_{t \to 0} \frac{x}{\sqrt{9 - x} - 3} \]
To simplify this expression, we multiply both the numerator and the denominator by the conjugate of the denominator:
\[ \frac{x}{\sqrt{9 - x} - 3} \times \frac{\sqrt{9 - x} + 3}{\sqrt{9 - x} + 3} = \frac{x(\sqrt{9 - x} + 3)}{(\sqrt{9 - x})^2 - 3^2} \]
The denominator simplifies as follows:
\[ (\sqrt{9 - x})^2 - 3^2 = 9 - x - 9 = -x \]
So the expression becomes:
\[ \frac{x(\sqrt{9 - x} + 3)}{-x} = -(\sqrt{9 - x} + 3) \]
Now, as \( x \to 0 \), we substitute \( x = 0 \) into the simplified expression:
\[ -(\sqrt{9 - 0} + 3) = -(3 + 3) = -6 \]
Answer: -6
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