Q.Find $\lim_{x\to 1} f(x)$, where $f(x) = \begin{cases} x^2 - 1, & x \le 1 \\ -x^2 - 1, & x > 1 \end{cases}$

The key idea is that for a piecewise function, the limit exists only if the left-hand and right-hand limits are equal.

Step 1 — Left-hand limit ($x \to 1^-$):

For $x \le 1$, $f(x) = x^2 - 1$.

$\displaystyle \lim_{x \to 1^-} (x^2 - 1) = 1^2 - 1 = 0$.

Step 2 — Right-hand limit ($x \to 1^+$):

For $x > 1$, $f(x) = -x^2 - 1$.

$\displaystyle \lim_{x \to 1^+} (-x^2 - 1) = -1^2 - 1 = -2$. …

Why we check both sides

When a function is defined by different expressions on either side of a point, the overall limit $\lim_{x\to 1} f(x)$ exists only if the function approaches the same value from the left and from the right. This is the core idea: the two one-sided limits must agree.

Here, $f(x)$ uses $x^2-1$ for $x\le 1$ and $-x^2-1$ for $x>1$. The break is at $x=1$, so we compute each side separately.

Step-by-step

1. Left-hand limit ($x\to 1^-$): For $x<1$, $f(x)=x^2-1$. This is a polynomial, so we can substitute directly:

2. Right-hand limit ($x\to 1^+$): For $x>1$, $f(x)=-x^2-1$. Again, substitute:

3. Compare the two: …