If `f(x)={{:((sqrt(x^(2)+5)-3)/(x+2)",",x ne-2),(k,","x=-2):}` is continuous at x=-2 , then the value of k is

Options

A. `-(2)/(3)`
B. 0
C. `(2)/(3)`
D. none of these

Answer

To determine the value of \( k \) for the function

\[ f(x) = \begin{cases} \frac{\sqrt{x^2 + 5} - 3}{x + 2} & \text{if } x \neq -2 \\ k & \text{if } x = -2 \end{cases} \]

to be continuous at \( x = -2 \), we need to ensure that

\[ f(-2) = \lim_{x \to -2} f(x). \]

### Step 1: Find \( f(-2) \)

Since \( f(-2) = k \), we have:

\[ f(-2) = k. \]

### Step 2: Find \( \lim_{x \to -2} f(x) \)

We need to calculate the limit:

\[ \lim_{x \to -2} \frac{\sqrt{x^2 + 5} - 3}{x + 2}. \]

### Step 3: Substitute \( x = -2 \)

Substituting \( x = -2 \) directly into the limit gives us:

\[ \frac{\sqrt{(-2)^2 + 5} - 3}{-2 + 2} = \frac{\sqrt{4 + 5} - 3}{0} = \frac{\sqrt{9} - 3}{0} = \frac{3 - 3}{0} = \frac{0}{0}. \]

This is an indeterminate form, so we need to manipulate the expression.

### Step 4: Rationalize the numerator

To resolve the indeterminate form, we multiply the numerator and denominator by the conjugate of the numerator:

\[ \lim_{x \to -2} \frac{\sqrt{x^2 + 5} - 3}{x + 2} \cdot \frac{\sqrt{x^2 + 5} + 3}{\sqrt{x^2 + 5} + 3}. \]

This gives us:

\[ \lim_{x \to -2} \frac{(\sqrt{x^2 + 5})^2 - 3^2}{(x + 2)(\sqrt{x^2 + 5} + 3)} = \lim_{x \to -2} \frac{x^2 + 5 - 9}{(x + 2)(\sqrt{x^2 + 5} + 3)} = \lim_{x \to -2} \frac{x^2 - 4}{(x + 2)(\sqrt{x^2 + 5} + 3)}. \]

### Step 5: Factor the numerator

The numerator \( x^2 - 4 \) can be factored as:

\[ x^2 - 4 = (x - 2)(x + 2). \]

Thus, we have:

\[ \lim_{x \to -2} \frac{(x - 2)(x + 2)}{(x + 2)(\sqrt{x^2 + 5} + 3)}. \]

### Step 6: Cancel \( (x + 2) \)

We can cancel \( (x + 2) \) from the numerator and denominator (as long as \( x \neq -2 \)):

\[ \lim_{x \to -2} \frac{x - 2}{\sqrt{x^2 + 5} + 3}. \]

### Step 7: Substitute \( x = -2 \) again

Now substituting \( x = -2 \):

\[ \frac{-2 - 2}{\sqrt{(-2)^2 + 5} + 3} = \frac{-4}{\sqrt{4 + 5} + 3} = \frac{-4}{\sqrt{9} + 3} = \frac{-4}{3 + 3} = \frac{-4}{6} = -\frac{2}{3}. \]

### Step 8: Set the limit equal to \( k \)

Since \( f \) is continuous at \( x = -2 \):

\[ k = \lim_{x \to -2} f(x) = -\frac{2}{3}. \]

### Final Answer

Thus, the value of \( k \) is:

\[ \boxed{-\frac{2}{3}}. \]