Evaluate : `("lim")_(xvec2a^+)(sqrt(x-2a)+sqrt(x)-sqrt(2a))/sqrt(x^2-4a^2)`

Options

A. `1/(sqrt(a))`
B. `1/(2sqrt(a))`
C. `(sqrt(a))/2`
D. `2sqrt(a)`

Answer

To evaluate the limit

\[ \lim_{x \to 2a^+} \frac{\sqrt{x - 2a} + \sqrt{x} - \sqrt{2a}}{\sqrt{x^2 - 4a^2}}, \]

we first substitute \(x = 2a\):

1. **Substituting \(x = 2a\)**:
\[ \sqrt{2a - 2a} + \sqrt{2a} - \sqrt{2a} = 0 + \sqrt{2a} - \sqrt{2a} = 0, \]
and for the denominator,
\[ \sqrt{(2a)^2 - 4a^2} = \sqrt{4a^2 - 4a^2} = \sqrt{0} = 0. \]
This gives us the indeterminate form \( \frac{0}{0} \).

2. **Applying L'Hôpital's Rule**:
Since we have the indeterminate form, we can apply L'Hôpital's Rule, which states that we can take the derivative of the numerator and the denominator separately.

- **Numerator**:
\[ \frac{d}{dx}(\sqrt{x - 2a} + \sqrt{x} - \sqrt{2a}) = \frac{1}{2\sqrt{x - 2a}} + \frac{1}{2\sqrt{x}}. \]
- **Denominator**:
\[ \frac{d}{dx}(\sqrt{x^2 - 4a^2}) = \frac{1}{2\sqrt{x^2 - 4a^2}} \cdot (2x) = \frac{x}{\sqrt{x^2 - 4a^2}}. \]

3. **Rewriting the limit**:
Now we rewrite our limit using the derivatives:
\[ \lim_{x \to 2a^+} \frac{\frac{1}{2\sqrt{x - 2a}} + \frac{1}{2\sqrt{x}}}{\frac{x}{\sqrt{x^2 - 4a^2}}}. \]

4. **Finding a common denominator**:
The numerator can be combined:
\[ \frac{1}{2\sqrt{x - 2a}} + \frac{1}{2\sqrt{x}} = \frac{\sqrt{x} + \sqrt{x - 2a}}{2\sqrt{x}\sqrt{x - 2a}}. \]
Thus, the limit becomes:
\[ \lim_{x \to 2a^+} \frac{\sqrt{x} + \sqrt{x - 2a}}{2\sqrt{x}\sqrt{x - 2a}} \cdot \frac{\sqrt{x^2 - 4a^2}}{x}. \]

5. **Simplifying the limit**:
We know that \(x^2 - 4a^2 = (x - 2a)(x + 2a)\). Thus,
\[ \sqrt{x^2 - 4a^2} = \sqrt{(x - 2a)(x + 2a)}. \]
The limit now looks like:
\[ \lim_{x \to 2a^+} \frac{\sqrt{x} + \sqrt{x - 2a}}{2\sqrt{x}\sqrt{x - 2a}} \cdot \frac{\sqrt{(x - 2a)(x + 2a)}}{x}. \]

6. **Canceling terms**:
The \(\sqrt{x - 2a}\) in the numerator and denominator cancels:
\[ \lim_{x \to 2a^+} \frac{\sqrt{x} + \sqrt{x - 2a}}{2\sqrt{x}} \cdot \frac{\sqrt{x + 2a}}{x}. \]

7. **Evaluating the limit**:
Substitute \(x = 2a\):
\[ \frac{\sqrt{2a} + 0}{2\sqrt{2a}} \cdot \frac{\sqrt{4a}}{2a} = \frac{\sqrt{2a}}{2\sqrt{2a}} \cdot \frac{2\sqrt{a}}{2a} = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2\sqrt{2}}. \]

Thus, the final answer is:

\[ \frac{1}{2\sqrt{2}}. \]