Continuity at a point AB & BC

$f$ is continuous at $x = a$ exactly when all three hold:

1. $f(a)$ is defined.
2. $\lim_{x\to a} f(x)$ exists, which itself requires both one-sided limits to exist and agree.
3. $\lim_{x\to a} f(x) = f(a)$: the approach and the arrival match.

Compactly, all of it is the single statement $\lim_{x\to a} f(x) = f(a)$, since writing that presupposes both sides exist. Expanded into three parts it is easier to check and easier to see what failed.

Condition 3 is the one that gets skipped, because 1 and 2 are the ones you naturally compute. Having found both, comparing them takes a second and is the entire content of the definition.

Three functions, each failing exactly one condition at $x = 2$:

1. $f(x) = \frac{x^2-4}{x-2}$. The limit is 4, but $f(2)$ is undefined. Condition 1 fails.
2. $f(x) = \frac{|x-2|}{x-2}$ with $f(2) = 0$. The value exists, but the one-sided limits are $-1$ and 1. Condition 2 fails.
3. $f(x) = x + 2$ for $x \ne 2$, with $f(2) = 99$. Both exist: the limit is 4 and the value is 99. Condition 3 fails.

The third case is the one to keep in mind. Everything a student would normally compute comes out fine, and the function is still discontinuous.

A standard exam task: find $k$ making a piecewise function continuous. Suppose $f(x) = x^2$ for $x < 3$ and $f(x) = kx - 3$ for $x \ge 3$.

1. Left limit: $\lim_{x\to 3^-} x^2 = 9$.
2. Right limit: $\lim_{x\to 3^+}(kx-3) = 3k-3$.
3. Value: $f(3) = 3k-3$, since the second rule owns $x = 3$.
4. Continuity needs all three equal: $9 = 3k-3$, so $k = 4$.

Step 3 is not optional even though it duplicates step 2 here. When the branch containing $x=a$ is not the one used for the matching one-sided limit, the value is a genuinely separate check, and skipping it can leave a function whose sides agree while the value sits somewhere else entirely.

Rather than testing point by point, it helps to know the standard continuous families. Polynomials are continuous on all of $\mathbb{R}$. Rational functions, radicals, exponentials, logarithms, and trigonometric functions are continuous on their domains, and that qualifier is where the exceptions live: $\frac{1}{x}$ is continuous at every point of its domain, and 0 is simply not in it.

Sums, products, quotients (with nonzero denominators), and compositions of continuous functions are continuous. So the interesting points are always the few where something is undefined or where a piecewise definition changes rule, and those are the only places worth testing.