Continuity

The three-part definition. A function $f$ is continuous at a point $x=a$ when all three quantities exist and coincide:

where: the left side is the left-hand limit (approach from values less than $a$), the middle is the right-hand limit (approach from values greater than $a$), and $f(a)$ is the function's actual value at $a$. If any one of the three fails to exist, or the three do not all match, $f$ is discontinuous at $a$. A function is continuous on an interval if it is continuous at every point of that interval; at the endpoints of a closed interval only the one-sided limit lying inside the interval is required.

Locality is the whole point. Continuity is defined pointwise, at a single value $a$, before being extended to a whole interval by demanding it hold everywhere in that interval. That is why a suspicious-looking single point — a seam in a piecewise definition, an integer for $[x]$, the origin for $|x|$ — is almost always where the real content of a continuity question lives; away from such points, most "ordinary" functions behave exactly as expected.

Algebra of continuous functions. If $f$ and $g$ are both continuous at $a$, then so are $f+g$, $f-g$, $f\cdot g$, and $f/g$ (provided $g(a)\ne 0$). Polynomials, and $\sin x$, $\cos x$, $e^x$, $\ln x$ on their natural domains, are continuous everywhere they are defined. Once you accept these building blocks, you can certify continuity of most combinations instantly, without computing a single limit — reserving the three-part definition for the specific points where a function is pieced together or built from $|x|$/$[x]$.

Removable discontinuity. This occurs when $\lim_{x\to a} f(x)$ exists (the two one-sided limits agree) but either $f(a)$ is undefined or $f(a)$ differs from that common limit. Redefining $f(a)$ to equal the limit "removes" the discontinuity — a distinction the figure below makes visual.

Examine the continuity of $f(x) = \dfrac{x^2-4}{x-2}$ for $x\ne 2$, with $f(2)=5$, at $x=2$; then discuss $g(x) = |x-3| + [x]$ at $x=3$.

Step 1 — left-hand limit of $f$. For $x\ne2$, $\dfrac{x^2-4}{x-2} = \dfrac{(x-2)(x+2)}{x-2} = x+2$, so $\lim_{x\to2^-} f(x) = \lim_{x\to2^-}(x+2) = 4$.

Step 2 — right-hand limit of $f$. By the same simplification, $\lim_{x\to2^+} f(x) = \lim_{x\to2^+}(x+2) = 4$. The two one-sided limits agree, so $\lim_{x\to2} f(x) = 4$.

Step 3 — compare to $f(2)$. We are given $f(2) = 5$, but the limit is $4$, so $f(2) \ne \lim_{x\to2} f(x)$. Since the limit exists but does not match the function value, $f$ has a removable discontinuity at $x=2$; redefining $f(2)=4$ would fix it — this is exactly the picture in the figure above.

Step 4 — the piecewise function $g$. Near $x=3$: for $x$ slightly less than $3$, $|x-3| = 3-x$ and $[x]=2$, so $\lim_{x\to3^-} g(x) = (3-3)+2 = 2$. For $x$ slightly more than $3$, $|x-3| = x-3$ and $[x]=3$, so $\lim_{x\to3^+} g(x) = (3-3)+3 = 3$. The one-sided limits disagree ($2 \ne 3$), so $\lim_{x\to3} g(x)$ does not exist and $g$ is discontinuous at $x=3$ — and because the two one-sided limits themselves differ, this discontinuity is not removable.

Final answers: $f$ has a removable discontinuity at $x=2$ (limit $4$, value $5$); $g$ has a non-removable discontinuity at $x=3$ (left limit $2$, right limit $3$), caused entirely by the jump in $[x]$.

Competitive stretch

If a third function is defined as $h(x) = \dfrac{x^2-4}{x-2}$ for $x\ne2$ and $h(2)=4$, then by Step 1–2 above, $\lim_{x\to2}h(x)=4=h(2)$, so $h$ is continuous at $x=2$ — the only difference from $f$ is the assigned value at the single point $x=2$, confirming that continuity can hinge entirely on one number.

• Confusing "the limit exists" with "the function is continuous" — a removable discontinuity has a perfectly good limit but is still discontinuous unless $f(a)$ matches it.
• Checking only one side of a piecewise function or a function built from $|x|$ or $[x]$; a false sense of continuity comes from evaluating just the left-hand or just the right-hand limit.
• Assuming $[x]$ is "continuous almost everywhere so it's fine" — the syllabus requires you to pinpoint the exact points (every integer) where it fails, not wave at general behaviour.
• Treating a closed-interval endpoint like an interior point — only the relevant one-sided limit is required there, not both.
• Assuming a function built from "nice" pieces (polynomials, trig functions) is automatically continuous everywhere those pieces are defined without checking the seams — the algebra-of-continuous-functions rule only guarantees continuity where the formula does not change.

Ask the tutor: Stuck on whether a discontinuity in your own piecewise function is removable, or want another $|x|$/$[x]$ example worked through step by step — ask the AI tutor.