Q.Find all points of discontinuity of $f$, where $f$ is defined by $f(x) = \begin{cases} \frac{x}{|x|}, & \text{if } x < 0 \\ -1, & \text{if } x \geq 0 \end{cases}$

Concept: Continuity At A Point — a function is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$.

We check continuity at $x = 0$, the only possible point of trouble.

1. For $x < 0$, $|x| = -x$, so $\frac{x}{|x|} = \frac{x}{-x} = -1$. Hence $\lim_{x \to 0^-} f(x) = -1$.

2. For $x \geq 0$, $f(x) = -1$, so $\lim_{x \to 0^+} f(x) = -1$.

3. Both one-sided limits equal $-1$, so $\lim_{x \to 0} f(x) = -1$. Also $f(0) = -1$. …

We are checking continuity of a piecewise function. The key idea: continuity at a point means the left-hand limit, right-hand limit, and the function's value at that point all agree. For a piecewise function, the only place where things can go wrong is at the boundary between the pieces — here, at $x = 0$. Everywhere else, each piece is a constant function, and constants are always continuous.

Let’s examine the definition carefully.

1. For $x < 0$: The rule says $f(x) = \frac{x}{|x|}$. Since $x$ is negative, $|x| = -x$. So $\frac{x}{|x|} = \frac{x}{-x} = -1$. That means for every negative $x$, $f(x) = -1$. So on $(-\infty, 0)$, $f$ is the constant function $-1$.

2. For $x \geq 0$: The rule directly says $f(x) = -1$. So on $[0, \infty)$, $f$ is also the constant function $-1$.

So the function is actually the same constant $-1$ everywhere — there is no break. But we must still formally check continuity at $x = 0$, because the definition changes there.

3. Check continuity at $x = 0$:
   • Left-hand limit: As $x \to 0^-$, we use the $x<0$ piece: $f(x) = -1$. So $\lim_{x \to 0^-} f(x) = -1$.
   • Right-hand limit: As $x \to 0^+$, we use the $x \geq 0$ piece: $f(x) = -1$. So $\lim_{x \to 0^+} f(x) = -1$.
   • Function value: $f(0) = -1$ (from the $x \geq 0$ rule).
   • Since left limit = right limit = $f(0)$, the function is continuous at $x = 0$. …

Method: Simplifying a Domain-Restricted Absolute-Value Expression Before Testing Continuity

Some piecewise definitions apply $|x|$ only on a restricted domain (e.g. only for $x<0$) rather than across all of $\mathbb{R}$ — this method shows how to handle that safely.

Steps

Step 1: Simplify $|x|$ only within the domain it is actually used on

If the formula uses $|x|$ for, say, $x<0$ only, substitute $|x|=-x$ (the rule valid on that restricted domain) rather than writing out the full two-branch definition of $|x|$ unnecessarily:

Step 2: Reduce each branch to its simplest form

After substitution, check whether the branch becomes a genuinely constant expression (as often happens when $x$ appears both in the numerator and denominator with matching sign). A branch that simplifies to a constant is trivially continuous on its own open interval.

Step 3: Test the boundary where the branches meet …

Common Mistakes

Mistake 1: Seeing an absolute value in the formula and assuming the function must be discontinuous somewhere

Why it's wrong: the presence of $|x|$ does not automatically create a jump — here $\frac{x}{|x|}$ for $x<0$ simplifies cleanly to the constant $-1$, with no sign change or blow-up anywhere on that restricted domain. Correct approach: always simplify the expression fully before deciding anything about its behaviour.

Mistake 2: Stopping after the two one-sided limits match, without checking $f(0)$ too …