`lim_(x rarr2)(5)/(sqrt(2)-sqrt(x))`

Options

A. `10sqrt(2)`
B. `oo`
C. `-oo`
D. non -existant

Explanation

To solve the limit \( \lim_{x \to 2} \frac{5}{\sqrt{2} - \sqrt{x}} \), we will evaluate both the left-hand limit (LHL) and the right-hand limit (RHL) to determine if the limit exists.

### Step 1: Evaluate the Left-Hand Limit (LHL)
The left-hand limit is evaluated as \( x \) approaches 2 from the left, which we denote as \( x \to 2^{-} \).

\[ \text{LHL} = \lim_{x \to 2^{-}} \frac{5}{\sqrt{2} - \sqrt{x}} \]

As \( x \) approaches 2 from the left, \( \sqrt{x} \) approaches \( \sqrt{2} \). Thus, \( \sqrt{2} - \sqrt{x} \) approaches \( 0 \) from the positive side, making the denominator approach \( 0^{+} \).

\[ \text{LHL} = \frac{5}{0^{+}} = +\infty \]

### Step 2: Evaluate the Right-Hand Limit (RHL)
The right-hand limit is evaluated as \( x \) approaches 2 from the right, denoted as \( x \to 2^{+} \).

\[ \text{RHL} = \lim_{x \to 2^{+}} \frac{5}{\sqrt{2} - \sqrt{x}} \]

As \( x \) approaches 2 from the right, \( \sqrt{x} \) approaches \( \sqrt{2} \). Thus, \( \sqrt{2} - \sqrt{x} \) approaches \( 0 \) from the negative side, making the denominator approach \( 0^{-} \).

\[ \text{RHL} = \frac{5}{0^{-}} = -\infty \]

### Step 3: Conclusion
Since the left-hand limit and the right-hand limit are not equal:

\[ \text{LHL} = +\infty \quad \text{and} \quad \text{RHL} = -\infty \]

The limit does not exist.

\[ \lim_{x \to 2} \frac{5}{\sqrt{2} - \sqrt{x}} \text{ does not exist.} \]

### Final Answer
The limit does not exist.

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