Q.Let $f(x) = \begin{cases} x + 2 & x \leq 1 \\ cx^2 & x > -1 \end{cases}$, find '$c$' if $\lim_{x \to -1} f(x)$ exists.

The key idea is that for the limit to exist at $x = -1$, the left-hand and right-hand limits must be equal. Since the function definition changes at $x = -1$, we evaluate each side separately.

Step 1: Left-hand limit ($x \to -1^-$, so $x \leq -1$).

Here $f(x) = x + 2$, so

Step 2: Right-hand limit ($x \to -1^+$, so $x > -1$). …

The key idea here is that a limit exists at a point only when the function approaches the same value from both sides. For a piecewise function, this means the expressions on either side of the boundary must agree at the boundary point.

Let’s unpack why this matters. The function $f(x)$ is defined differently for $x \leq 1$ and for $x > -1$. Notice the overlap: both pieces cover the region around $x = -1$. The first piece, $x+2$, applies for all $x \leq 1$ (which includes $x = -1$). The second piece, $cx^2$, applies for all $x > -1$ (which also includes values just to the right of $-1$, but not $-1$ itself). So at $x = -1$, the function is defined by the first piece, but the limit from the right depends on the second piece.

For $\lim_{x \to -1} f(x)$ to exist, the left-hand limit and right-hand limit must be equal. Let’s compute them.

1. Left-hand limit ($x \to -1^-$): When $x$ approaches $-1$ from the left, $x < -1$, so we use the first piece: $f(x) = x + 2$.

2. Right-hand limit ($x \to -1^+$): When $x$ approaches $-1$ from the right, $x > -1$, so we use the second piece: $f(x) = cx^2$.

3. Equate the two limits: For the overall limit to exist, $\lim_{x \to -1^-} f(x) = \lim_{x \to -1^+} f(x) \quad \Rightarrow \quad 1 = c.$ …