Let `f(x)={{:(x+k,,,x le -1), (cx^(2),,, x gt -1","):}` find 'c', if `lim_(x to -1)f(x)` exists.

Answer

To solve the problem, we need to find the value of \( c \) such that the limit of the function \( f(x) \) exists as \( x \) approaches \(-1\). The function is defined as follows:

\[ f(x) = \begin{cases} x + k & \text{if } x \leq -1 \\ c x^2 & \text{if } x > -1 \end{cases} \]

### Step 1: Find the Left-Hand Limit (LHL)

To find the left-hand limit as \( x \) approaches \(-1\), we use the part of the function defined for \( x \leq -1 \):

\[ \text{LHL} = \lim_{x \to -1^-} f(x) = \lim_{x \to -1^-} (x + k) \]

Substituting \( x = -1 \):

\[ \text{LHL} = -1 + k = k - 1 \]

### Step 2: Find the Right-Hand Limit (RHL)

Next, we find the right-hand limit as \( x \) approaches \(-1\), using the part of the function defined for \( x > -1 \):

\[ \text{RHL} = \lim_{x \to -1^+} f(x) = \lim_{x \to -1^+} (c x^2) \]

Substituting \( x = -1 \):

\[ \text{RHL} = c \cdot (-1)^2 = c \]

### Step 3: Set LHL Equal to RHL

For the limit to exist at \( x = -1 \), the left-hand limit must equal the right-hand limit:

\[ \text{LHL} = \text{RHL} \]

Thus, we have:

\[ k - 1 = c \]

### Step 4: Conclusion

The value of \( c \) is expressed in terms of \( k \):

\[ c = k - 1 \]

This is the final answer.