Question

\(\lim\limits_{t\rightarrow0}\frac{x}{\sqrt{9-x}-3}\) is equal to

• A. 6
• B. 3
• C. -3
• D. -6
• E. 0

Answer 1

The Correct Option is D

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