Continuity of functions

A function $f$ is continuous at $x=a$ in its domain if the limit of $f$ as $x\to a$ exists and equals the function's value there:

where: $f(a)$ is the value the function actually takes at $a$, and $\lim_{x\to a}f(x)$ is the number the function's outputs approach as $x$ gets arbitrarily close to $a$ from both sides. This one line hides three separate conditions, and continuity fails if even one of them breaks.

Condition 1 — defined. $f(a)$ must exist; the point must genuinely sit in the domain.

Condition 2 — limit exists. The left-hand limit (LHL) and right-hand limit (RHL) must be equal:

where: the superscripts $a^-$ and $a^+$ mean $x$ approaches $a$ from below and from above respectively.

Condition 3 — they match the value. The common limit from Condition 2 must equal $f(a)$ from Condition 1. Standard building blocks — polynomials, $\sin x$, $\cos x$, $e^x$, and (on their domains) $\tan x$, $\log x$, rational functions — are continuous everywhere they are defined, so the interesting work concentrates at a small set of "suspicious" points: where a piecewise formula switches, where a denominator could vanish, or where a step function like the greatest integer function $[x]$ jumps. A function is continuous on an open interval $(a,b)$ if it passes the test at every point inside it; on a closed interval $[a,b]$ it must additionally be right-continuous at $a$ (i.e. $\lim_{x\to a^+}f(x)=f(a)$) and left-continuous at $b$ (i.e. $\lim_{x\to b^-}f(x)=f(b)$) — no new definition, just the point test applied repeatedly, with the endpoints only required to behave from the one side that actually lies inside the interval. It also helps to think of continuity as a local property first, then a global one. "Continuous on an interval" is not a separate rule to memorise — it is simply the point-wise test succeeding at every single point of that interval, one point at a time. This is why an exam question rarely asks you to check continuity everywhere from scratch; instead it hands you a formula built from continuous pieces and asks you to hunt down the handful of places where the formula could plausibly break — a join, a vanishing denominator, or an integer where a floor function steps.

The goal is to show, using nothing beyond the three-part test and basic algebra, that requiring a piecewise function to be continuous at its join point turns into a single solvable equation for an unknown constant — a task that appears repeatedly in board exams whenever a function is defined by two different formulas and asked to be continuous at the switch-over point.

Step 1. Set up the function. Let $f(x)=kx+3$ for $x\leq 2$ and $f(x)=2x-5$ for $x>2$, where $k$ is an unknown constant. Since $x=2$ is where the formula switches, this is the only point that needs checking; each piece is a polynomial, hence automatically continuous everywhere on its own side.

Step 2. Compute the value at $x=2$ using the piece that applies there. Because $2\leq 2$, use the first piece: $f(2) = k(2)+3 = 2k+3$.

Step 3. Compute the left-hand limit using the first piece (it governs $x<2$ as well): $\lim_{x\to 2^-} f(x) = k(2)+3 = 2k+3$. This equals $f(2)$ automatically, since the same formula supplies both.

Step 4. Compute the right-hand limit using the second piece: $\lim_{x\to 2^+} f(x) = 2(2)-5 = -1$.

Step 5. Apply the continuity condition, LHL $=$ RHL: $2k+3 = -1$, so $2k=-4$ and $k=-2$. Only this value of $k$ makes the graph join without a jump; any other value leaves a gap exactly at $x=2$.

Check whether $f(x)=\dfrac{x^2-4}{x-2}$ for $x\neq 2$, with $f(2)=4$, is continuous at $x=2$.

Step 1. Read off $f(2)$ directly from the definition: $f(2)=4$.

Step 2. Simplify the formula for $x\neq 2$: $f(x)=\dfrac{x^2-4}{x-2}=\dfrac{(x-2)(x+2)}{x-2}=x+2$, valid because $x\neq 2$ lets the common factor cancel.

Step 3. Take the limit: $\lim_{x\to 2} f(x) = \lim_{x\to 2}(x+2) = 4$.

Step 4. Compare: $\lim_{x\to 2} f(x) = 4 = f(2)$, so all three conditions hold and $f$ is continuous at $x=2$. Without the extra definition $f(2)=4$, the expression would have had a hole at $x=2$; defining $f(2)=4$ precisely removes that removable discontinuity.

Competitive stretch

The greatest integer function $f(x)=[x]$ is a useful contrast: at $x=2$, $\text{LHL}=\lim_{x\to 2^-}[x]=1$ (values just below $2$ round down to $1$) while $\text{RHL}=\lim_{x\to 2^+}[x]=2$ and $f(2)=2$. Since LHL $\neq$ RHL, $[x]$ is discontinuous at every integer, even though it is defined everywhere — a sharp reminder that "defined" and "continuous" are not the same thing.

• Confusing "defined" with "continuous." A jump function can be defined at every point and still fail continuity, because the limit must match the value, not just exist as a number to substitute.
• Checking only one side. For piecewise functions, students often compute the right-hand limit (or substitute directly) and skip the left-hand limit, missing a mismatch at the join.
• Cancelling factors carelessly. In $\dfrac{x^2-4}{x-2}$, cancelling $(x-2)$ is valid only for $x\neq 2$; the simplified form $x+2$ describes the limiting behaviour, not automatically the value at $x=2$ unless separately defined.
• Assuming "nice" pieces guarantee continuity everywhere. Each piece being a polynomial does not excuse checking the join point explicitly — continuity elsewhere never guarantees continuity at the boundary.

Ask the tutor: Want to work through a few more piecewise continuity checks — including ones where you solve for two unknown constants at once — until the left-hand/right-hand limit habit feels automatic?