Let `f(x)={{:(x+2,,x le -1), (xc^(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 \( \lim_{x \to -1} f(x) \) exists. The function \( f(x) \) is defined as follows:

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

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

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

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

Substituting \( x = -1 \):

\[ = -1 + 2 = 1 \]

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

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

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

Substituting \( x = -1 \):

\[ = -1 \cdot c^2 = -c^2 \]

### Step 3: Set the Limits Equal

For the limit \( \lim_{x \to -1} f(x) \) to exist, the left-hand limit must equal the right-hand limit:

\[ 1 = -c^2 \]

### Step 4: Solve for \( c^2 \)

Rearranging the equation gives:

\[ c^2 = -1 \]

### Step 5: Analyze the Result

The equation \( c^2 = -1 \) has no real solutions because the square of a real number cannot be negative. Thus, there is no value of \( c \) that will make the limit exist.

### Conclusion

The final conclusion is that there is no value of \( c \) such that \( \lim_{x \to -1} f(x) \) exists.

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