Continuity across an interval AB & BC

$f$ is continuous on an interval $I$ when it is continuous at every point of $I$. Verifying it at $x = 0$ says nothing about $x = 1.5$, so a single successful test is not evidence for the interval.

Checking infinitely many points is impossible, so the argument runs the other way: name the family, then find the exceptions.

1. Identify what the function is built from. Polynomials are continuous everywhere; rational, radical, exponential, logarithmic, and trigonometric functions are continuous on their domains.
2. List the finitely many suspect points: zeros of denominators, endpoints of radicands, and piecewise joins.
3. Check each suspect. If none lies in $I$, or each passes, $f$ is continuous on $I$.

That converts an infinite verification into a short finite one.

On a closed interval $[a, b]$, continuity at the endpoints means one-sided continuity, since the function need not be defined outside:

$$\lim_{x\to a^+} f(x) = f(a), \qquad \lim_{x\to b^-} f(x) = f(b).$$

Demanding two-sided continuity at $a$ would be asking about inputs less than $a$, which are outside the interval and irrelevant to the claim.

This matters because $\sqrt{x}$ is continuous on $[0, \infty)$ even though it has no left-hand limit at 0. Rejecting it on those grounds is a real error, and it is the reason the Intermediate Value Theorem can be applied to functions like it on closed intervals starting at their domain's edge.

Three failures account for nearly all of them:

1. An asymptote inside the interval. $\frac{1}{x}$ is continuous on $[1, 5]$ and on $[-5, -1]$, but not on $[-1, 1]$, because 0 sits inside and the function is not even defined there.
2. A piecewise join inside the interval where the pieces fail to meet.
3. A domain edge inside the interval. $\sqrt{x-2}$ is not continuous on $[0, 5]$, since it is undefined on part of that interval.

The habit that catches all three: before answering, sketch or list what lies strictly inside the interval. A function can be beautifully behaved at both endpoints and undefined in the middle.

The bracket type changes the claim. $\frac{1}{x}$ is continuous on $(0, 1)$ and also on $(0, 1]$, but not on $[0, 1]$, because 0 is now included and the function has no value there.

Conversely, a function can be continuous on a closed interval while failing just outside it, which is fine: the claim only concerns the interval named. Reading the brackets carefully is part of answering the question, and exam questions frequently vary only the bracket between choices.