The value of `lim_(xto2a)(sqrt(x-2a)+sqrt(x)-sqrt(2a))/(sqrt(x^2-4a^2))`is

Options

A. `4/(sqrt(a))`
B. `2/(sqrt(a))`
C. `1/(sqrt(a))`
D. `1/(2sqrt(a))`

Answer

To solve the limit

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

we will follow these steps:

### Step 1: Substitute the limit
First, we substitute \(x = 2a\) into the expression to check if we get an indeterminate form:

\[ \sqrt{2a - 2a} + \sqrt{2a} - \sqrt{2a} = 0 + \sqrt{2a} - \sqrt{2a} = 0, \]

and for the denominator:

\[ \sqrt{(2a)^2 - 4a^2} = \sqrt{4a^2 - 4a^2} = 0. \]

Since we have a \( \frac{0}{0} \) indeterminate form, we need to simplify the expression.

**Hint:** Check for indeterminate forms by substituting the limit value.

### Step 2: Simplify the numerator
We will rationalize the numerator. The term \(\sqrt{x} - \sqrt{2a}\) can be rationalized:

\[ \sqrt{x} - \sqrt{2a} = \frac{(\sqrt{x} - \sqrt{2a})(\sqrt{x} + \sqrt{2a})}{\sqrt{x} + \sqrt{2a}} = \frac{x - 2a}{\sqrt{x} + \sqrt{2a}}. \]

Thus, we can rewrite the limit as:

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

**Hint:** Use rationalization to simplify expressions involving square roots.

### Step 3: Combine the terms in the numerator
Now, we can combine the terms in the numerator:

\[ \sqrt{x - 2a} + \frac{x - 2a}{\sqrt{x} + \sqrt{2a}} = \frac{(\sqrt{x - 2a})(\sqrt{x} + \sqrt{2a}) + (x - 2a)}{\sqrt{x} + \sqrt{2a}}. \]

This gives us:

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

**Hint:** Combine fractions when possible to simplify the limit.

### Step 4: Simplify the denominator
The denominator can be simplified using the difference of squares:

\[ \sqrt{x^2 - 4a^2} = \sqrt{(x - 2a)(x + 2a)}. \]

Thus, we rewrite the limit as:

\[ \lim_{x \to 2a} \frac{(\sqrt{x - 2a})(\sqrt{x} + \sqrt{2a}) + (x - 2a)}{(\sqrt{x} + \sqrt{2a}) \sqrt{(x - 2a)(x + 2a)}}. \]

**Hint:** Use algebraic identities to simplify expressions involving square roots.

### Step 5: Cancel common factors
Now, we can see that \((x - 2a)\) appears in both the numerator and denominator. We can cancel \(\sqrt{x - 2a}\):

\[ \lim_{x \to 2a} \frac{(\sqrt{x} + \sqrt{2a}) + 1}{\sqrt{x + 2a}}. \]

### Step 6: Substitute again
Now we substitute \(x = 2a\):

\[ \frac{(\sqrt{2a} + \sqrt{2a}) + 1}{\sqrt{4a}} = \frac{2\sqrt{2a} + 1}{2\sqrt{a}}. \]

### Step 7: Final simplification
This simplifies to:

\[ \frac{2\sqrt{2a} + 1}{2\sqrt{a}} = \frac{2\sqrt{2} + \frac{1}{\sqrt{a}}}{2}. \]

Thus, the limit evaluates to:

\[ \frac{1}{2} \cdot \left(2\sqrt{2} + 1\right). \]

**Final Answer:** The value of the limit is \( \frac{1}{2} \cdot (2\sqrt{2} + 1) \).