Q.Is the function $f$ defined by $f(x) = \begin{cases} x, & \text{if } x \leq 1 \\ 5, & \text{if } x > 1 \end{cases}$ continuous at $x = 0$? At $x = 1$? At $x = 2$?

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

At $x = 0$:

For $x \leq 1$, $f(x) = x$, so $f(0) = 0$ and $\lim_{x \to 0} f(x) = 0$. The limit equals the function value.

$\Rightarrow$ $f$ is continuous at $x = 0$.

At $x = 1$:

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

Right-hand limit: $\lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} 5 = 5$.

Since $1 \neq 5$, the two-sided limit does not exist. …

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: $\lim_{x \to a} f(x) = f(a)$.

If any of these fails, the function is discontinuous at that point. For a piecewise function like this one, the only place where trouble can arise is at the "seam" — the point where the definition changes. Here, that seam is at $x = 1$. At other points, the function is just a simple polynomial or a constant, both of which are continuous everywhere.

Let's check each point the question asks about.

1. At $x = 0$

At $x = 0$, the function is defined by the first piece: $f(x) = x$, since $0 \leq 1$.

• $f(0) = 0$.
• Near $x = 0$, the function is just the line $y = x$, which is continuous everywhere. So $\lim_{x \to 0} f(x) = 0$.
• Since $\lim_{x \to 0} f(x) = f(0)$, the function is continuous at $x = 0$.

2. At $x = 1$

This is the critical point. The function changes its rule here, so we must check the left-hand limit and right-hand limit separately.

• Left-hand limit ($x \to 1^-$): For $x$ just less than 1, $f(x) = x$. So

• Right-hand limit ($x \to 1^+$): For $x$ just greater than 1, $f(x) = 5$. So

Since the left-hand limit (1) and the right-hand limit (5) are not equal, the two-sided limit $\lim_{x \to 1} f(x)$ does not exist.

• Function value: $f(1) = 1$ (since $1 \leq 1$).

Even though $f(1)$ exists, the limit does not exist, so the function is discontinuous at $x = 1$. …

Method: Checking Continuity of a Piecewise Function at Several Given Points

When asked to check continuity 'at $x=a$, at $x=b$, at $x=c$' for a piecewise function, treat each point separately and first decide whether it is an interior point of one piece or a boundary/seam point where the rule changes.

Steps

Step 1: Classify each named point

For a point strictly inside one piece's interval, the function is simply that one formula near the point (a polynomial or standard function), so continuity there is automatic and needs only direct substitution.

Step 2: At a boundary/seam point, compute the function value using the rule that actually applies there

Read the inequality carefully (e.g. $x\le a$ vs $x>a$) to determine which piece's formula gives $f(a)$ — this is the value continuity will be measured against.

Step 3: Compute the left-hand and right-hand limits separately

$\lim_{x\to a^-} f(x) \quad\text{and}\quad \lim_{x\to a^+} f(x)$ …

Common Mistakes

Mistake 1: Doing a full left/right-limit check at every named point, including points far from any seam

Why it's wrong: it wastes time and invites arithmetic slips — at $x=0$ and $x=2$ in this problem, the function is governed by a single piece near each point, so a one-sided split is unnecessary. Correct approach: first identify where the piecewise definition actually changes (here, $x=1$), and only do the two-sided limit comparison there.

Mistake 2: Concluding continuity at the seam point just because $f(a)$ equals one of the one-sided limits …