Q.If $f(x) = \begin{cases} |x| + 1, & x < 0 \\ 0, & x = 0 \\ |x| - 1, & x > 0 \end{cases}$. For what value(s) of $a$ does $\lim_{x\to a} f(x)$ exist?

Concept: Limit Existence — Check both one-sided limits at the point.

For $a < 0$, near $a$ we use $f(x) = |x| + 1 = -x + 1$ (since $x<0$). This is a polynomial, so the limit exists for all $a<0$.

For $a > 0$, near $a$ we use $f(x) = |x| - 1 = x - 1$. Again a polynomial, so the limit exists for all $a>0$.

The only potential trouble is at $a = 0$. Compute:

• Left-hand limit: $\lim_{x\to 0^-} f(x) = \lim_{x\to 0^-} (-x + 1) = 1$. …

We need to find all real numbers $a$ such that $\lim_{x \to a} f(x)$ exists. The function is defined piecewise, so the key is to check behaviour around the point where the definition changes — that is, at $x = 0$. For any $a \neq 0$, the function is simply $|x| + 1$ or $|x| - 1$ in a small neighbourhood around $a$, and both are continuous there. The only potential trouble is at $a = 0$, where the left and right definitions differ.

Let’s go step by step.

1. Understand the function near any $a$

   For $x < 0$, $|x| = -x$, so $f(x) = -x + 1$.

   For $x > 0$, $|x| = x$, so $f(x) = x - 1$.

   At $x = 0$, $f(0) = 0$ (but the limit doesn’t care about the value at the point).

   So the function is:

2. Case 1: $a < 0$ If $a$ is negative, then for all $x$ sufficiently close to $a$ (but not equal to $a$), we are still in the region $x < 0$. So near $a$, $f(x) = -x + 1$, which is a linear (hence continuous) function. Therefore,

The limit exists for every $a < 0$.

3. Case 2: $a > 0$ Similarly, if $a$ is positive, then near $a$ we have $f(x) = x - 1$, continuous. So

The limit exists for every $a > 0$.

4. Case 3: $a = 0$ — the critical point Here we must check the left-hand limit and the right-hand limit separately.
   • Left-hand limit ($x \to 0^-$): For $x$ just less than 0, $f(x) = -x + 1$. As $x \to 0^-$, $-x \to 0^+$, so $$$$