1.4 Q-4

Question Statement

Find the value of $c$ such that the following piecewise function has a limit at $x = -1$:

$$f(x) = \begin{cases} x + 2, & x \leq -1 c + 2, & x > -1 \end{cases}$$

Background and Explanation

For the limit of $f(x)$ to exist at $x = -1$, the left-hand limit (L.H.L) and right-hand limit (R.H.L) must both exist and be equal:

$$\lim_{x \to -1^-} f(x) = \lim_{x \to -1^+} f(x)$$

Additionally, the limit value must match the function’s value if continuity is required, but in this case, the focus is only on the existence of the limit.

Solution

Step 1: Evaluate the Left-Hand Limit (L.H.L)

For $x \leq -1$, the function is defined as $f(x) = x + 2$. Substituting $x \to -1^-$:

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

Step 2: Evaluate the Right-Hand Limit (R.H.L)

For $x > -1$, the function is defined as $f(x) = c + 2$. Substituting $x \to -1^+$:

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

Step 3: Set L.H.L Equal to R.H.L

For the limit at $x = -1$ to exist:

$$\lim_{x \to -1^-} f(x) = \lim_{x \to -1^+} f(x)$$

Substituting the results:

$$1 = c + 2$$

Step 4: Solve for $c$

Rearranging:

$$c = 1 - 2 = -1$$

Final Answer

The value of $c$ must be $-1$ for the limit of $f(x)$ to exist at $x = -1$.

Key Formulas or Methods Used

1. Definition of Limit Existence:
   A limit at $x = a$ exists if $\lim_{x \to a^-} f(x) = \lim_{x \to a^+} f(x)$.

2. Piecewise Function Analysis:
   Evaluate the function separately for $x < a$ and $x > a$, then equate the results.

Summary of Steps

1. Identify the function for $x \leq -1$ ($f(x) = x + 2$) and compute $\lim_{x \to -1^-} f(x) = 1$.
2. Identify the function for $x > -1$ ($f(x) = c + 2$) and compute $\lim_{x \to -1^+} f(x) = c + 2$.
3. Set $\lim_{x \to -1^-} f(x) = \lim_{x \to -1^+} f(x)$, yielding $1 = c + 2$.
4. Solve for $c$, obtaining $c = -1$.