Q.Find all points of discontinuity of $f$, where $f(x) = \begin{cases} \frac{\sin x}{x}, & \text{if } x < 0 \\ x+1, & \text{if } x \ge 0 \end{cases}$

Concept: Continuity At A Point — a function is continuous at $x = a$ if $\lim_{x \to a} f(x) = f(a)$. For piecewise functions, check the meeting point of the pieces.

Step 1: The only possible point of discontinuity is $x = 0$, where the definition changes.

Step 2: Compute the left-hand limit:

Step 3: Compute the right-hand limit and the function value:

$\lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} (x+1) = 1, \quad f(0) = 0 + 1 = 1$ …

The Core Idea: Continuity at a Point

A function is continuous at a point $x = a$ if three things happen:

1. The function is defined at $a$ (i.e., $f(a)$ exists).
2. The limit of $f(x)$ as $x$ approaches $a$ exists.
3. That limit equals the function value: $\lim_{x \to a} f(x) = f(a)$.

For a piecewise function, the only place where things can go wrong is at the boundary where the definition changes — here, that's $x = 0$. Everywhere else ($x < 0$ and $x > 0$), the function is given by a single, well-behaved formula ($\frac{\sin x}{x}$ for $x<0$ and $x+1$ for $x>0$), so it's automatically continuous there. The entire question boils down to: what happens at $x = 0$?

Step-by-Step Check at $x = 0$

1. Check the function value at $x = 0$.

Since the definition says $f(x) = x+1$ when $x \ge 0$, we use that piece at $x=0$:

So the function is defined, and $f(0) = 1$.

2. Find the left-hand limit (as $x \to 0^-$).

For $x < 0$, $f(x) = \frac{\sin x}{x}$. We need:

This is the classic standard limit. As $x$ approaches $0$ from the left, $\frac{\sin x}{x}$ approaches $1$. (The sign of $x$ doesn't matter here — the limit is symmetric.)

3. Find the right-hand limit (as $x \to 0^+$).

For $x > 0$, $f(x) = x+1$. This is a straight line, so:

4. Compare all three.

We have:

• Left-hand limit = $1$
• Right-hand limit = $1$
• Function value $f(0) = 1$

Since $\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x) = f(0)$, the function is continuous at $x = 0$. …

Method: Locating All Points of Discontinuity of a Piecewise Function

Use this whenever asked to "find all points of discontinuity" of a function defined piecewise — the search is narrowed to the boundary point(s) only, since each individual piece (away from the boundary) is built from standard continuous expressions.

Steps

Step 1: Narrow the search to the boundary point(s)

Away from where the formula switches, each piece is a single continuous expression (rational, polynomial, trig, etc.), so continuity there is automatic within its own domain. Only the join point(s) need to be checked explicitly.

Step 2: Compute the left-hand limit using the piece valid on that side

If the piece contains a standard indeterminate-form limit (such as $\dfrac{\sin x}{x}$ as $x \to 0$), use the known standard result:

Step 3: Compute the right-hand limit and the function value using the appropriate piece

Substitute directly since the piece covering the boundary side is typically a standard continuous expression there.

Step 4: Compare all three quantities …

Common Mistakes

Mistake 1: Assuming a piecewise function must be discontinuous simply because its formula changes at a point

Why it's wrong: a change in formula only creates a potential trouble spot — the two pieces can still meet up perfectly smoothly, as they do here. Correct approach: always compute the actual one-sided limits and function value before concluding anything; never assume a break just from seeing a piecewise definition.

Mistake 2: Misapplying or misremembering the standard limit $\lim_{x\to0}\frac{\sin x}{x}=1$

Why it's wrong: some students think this limit only applies as $x \to 0^+$, or confuse it with $\lim_{x\to 0}\frac{x}{\sin x}$ — using the wrong standard result gives a wrong one-sided limit. Correct approach: memorize the exact statement (numerator and denominator both $x$, limit equals $1$ from either side since $\sin x/x$ is an even function) and verify the expression matches this form exactly before applying it. …