Continuity of Functions on R; Epsilon-Delta

At a Glance

• Frequency: 2 sub-parts across 2 of 14 years (2018, 2026)
• Priority tier: T4
• Marks (count): 15 (1)
• Average solve time: ~18 min
• Difficulty mix: medium 1
• Section: B | Dominant type: proof

Why This Chapter Matters

This atom appears once in 13 years (2018, 15 marks) and is rated T4, so it is not a priority-one topic. However, the $\varepsilon$-$\delta$ technique underpins every continuity argument in the paper, and UPSC’s question type is tightly predictable: prove continuity or discontinuity of a specific function from the definition. Mastering the template here pays dividends in related atoms (uniform continuity, P2-RA-09; properties on compact sets, P2-RA-10).

Minimum Theory

Definition (Continuity at a point). Let $f:\mathbb{R}\to\mathbb{R}$ and $x_0 \in \mathbb{R}$. Then $f$ is continuous at $x_0$ if

$\forall\,\varepsilon > 0,\;\exists\,\delta > 0 \text{ such that } |x - x_0| < \delta \;\Rightarrow\; |f(x) - f(x_0)| < \varepsilon.$

Sequential characterisation. $f$ is continuous at $x_0$ if and only if for every sequence $(x_n)$ with $x_n \to x_0$ we have $f(x_n) \to f(x_0)$.

Discontinuity via sequences. To prove $f$ is discontinuous at $x_0$, it suffices to exhibit a sequence $x_n \to x_0$ such that $f(x_n) \not\to f(x_0)$.

Algebra of continuous functions. If $f$ and $g$ are continuous at $x_0$, so are $f+g$, $fg$, and $f/g$ (provided $g(x_0) \neq 0$). Composition: if $g$ is continuous at $x_0$ and $f$ is continuous at $g(x_0)$, then $f \circ g$ is continuous at $x_0$.

Classifying a discontinuity. At a point $x_0$ where $f$ fails to be continuous:

• First kind (jump): both one-sided limits exist and are finite but disagree (or disagree with $f(x_0)$). The jump is $\bigl|f(x_0^+)-f(x_0^-)\bigr|$. If the two one-sided limits agree with each other but not with $f(x_0)$, the discontinuity is removable.
• Second kind: at least one one-sided limit fails to exist — typically by oscillation.

Naming which kind you have, and producing both one-sided limits, is what separates a full answer from “it is discontinuous there”.

Functions defined as a pointwise limit — the $t^n$ trichotomy. When $f$ is defined by $f(x)=\lim_{n\to\infty}g_n(x)$ with $g_n$ built from an $n$-th power, the limit is governed entirely by the base against $1$. For $g_n(t)=\dfrac{t^n-1}{t^n+1}$ with $t\ge0$:

$t>1\Rightarrow t^n\to\infty\Rightarrow g_n\to 1;\qquad t=1\Rightarrow g_n\equiv0;\qquad 0\le t<1\Rightarrow t^n\to0\Rightarrow g_n\to-1,$

so $f=\operatorname{sgn}(t-1)$. The three branches must all be taken, including the endpoint $t=0$ if the base can vanish. The resulting $f$ is a step function, and the question then reduces to locating where the base crosses $1$. No $\varepsilon$-$\delta$ work is involved — the limit is evaluated case-wise and the continuity verdict is read off the steps.

Key examples.

• $f(x) = x^n$ is continuous everywhere (polynomial).
• $f(x) = |x|$ is continuous everywhere; $\delta = \varepsilon$ works by the reverse triangle inequality.
• Dirichlet function $f(x) = \mathbf{1}_{\mathbb{Q}}(x)$ is discontinuous everywhere.
• $f(x) = x \sin(1/x)$ for $x \neq 0$, $f(0) = 0$: continuous at $0$ (squeeze theorem).

Proving continuity of $f(x) = x^2$ at $x_0$. For $|x - x_0| < \delta$,

$|f(x) - f(x_0)| = |x^2 - x_0^2| = |x + x_0|\,|x - x_0|.$

Restrict $\delta \leq 1$ so that $|x - x_0| < 1$ implies $|x| < |x_0| + 1$, giving $|x + x_0| \leq |x| + |x_0| < 2|x_0| + 1$. Then

$|f(x) - f(x_0)| < (2|x_0| + 1)\,\delta.$

Choose $\delta = \min\!\left(1,\, \dfrac{\varepsilon}{2|x_0|+1}\right)$. Then $|f(x) - f(x_0)| < \varepsilon$.

Question Archetypes

prove-continuity (1 question(s); 2018)

Recognition Cues

• A specific function is named (polynomial, $|x|$, $\sqrt{x}$, $x\sin(1/x)$, etc.).
• Asked to verify continuity using the $\varepsilon$-$\delta$ definition.

Solution Template

1. Write out $|f(x) - f(x_0)|$ explicitly.
2. Factor or bound the expression to isolate $|x - x_0|$.
3. Impose a preliminary bound $\delta \leq 1$ (or another convenient constant) to control the remaining factor.
4. Choose $\delta = \min(1, \varepsilon/C)$ where $C$ is the bound on the other factor.
5. Verify: $|x - x_0| < \delta \Rightarrow |f(x) - f(x_0)| < C \cdot \delta \leq \varepsilon$.

Worked Example

2018 Paper 2, 2018-P2-Q2a (15 marks)

Using the $\varepsilon$-$\delta$ definition, prove that $f(x) = x^2$ is continuous at every point $x_0 \in \mathbb{R}$.

Proof.

Let $x_0 \in \mathbb{R}$ and $\varepsilon > 0$ be given. We need to find $\delta > 0$ such that

$|x - x_0| < \delta \;\Longrightarrow\; |x^2 - x_0^2| < \varepsilon.$

Observe that

$|x^2 - x_0^2| = |x - x_0|\,|x + x_0|.$

Preliminary bound. Assume first that $\delta \leq 1$, so $|x - x_0| < 1$. Then

$|x| \leq |x - x_0| + |x_0| < 1 + |x_0|,$

and by the triangle inequality,

$|x + x_0| \leq |x| + |x_0| < (1 + |x_0|) + |x_0| = 1 + 2|x_0|.$

Choice of $\delta$. Set

$\delta = \min\!\left(1,\; \frac{\varepsilon}{1 + 2|x_0|}\right).$

Verification. For $|x - x_0| < \delta$,

$|x^2 - x_0^2| = |x - x_0|\,|x + x_0| < \delta\,(1 + 2|x_0|) \leq \frac{\varepsilon}{1+2|x_0|}\,(1 + 2|x_0|) = \varepsilon.$

Since $\varepsilon > 0$ was arbitrary and $\delta$ depends only on $\varepsilon$ and $x_0$, $f$ is continuous at $x_0$. As $x_0$ was arbitrary, $f(x) = x^2$ is continuous on $\mathbb{R}$.

$\boxed{f(x) = x^2 \text{ is continuous on } \mathbb{R}.}\;\blacksquare$

Common Traps

• Choosing $\delta$ without first imposing a preliminary bound; this leaves the factor $|x + x_0|$ unbounded and the proof invalid.
• Writing $\delta = \varepsilon/(1 + 2|x_0|)$ without the $\min$ with $1$; then the preliminary bound $|x| < 1 + |x_0|$ need not hold.
• Confusing pointwise continuity (here, $\delta$ may depend on $x_0$) with uniform continuity (where $\delta$ depends only on $\varepsilon$). For $f(x) = x^2$, the dependence on $x_0$ is essential — uniform continuity fails on all of $\mathbb{R}$.

limit-function-discontinuities (1 question(s); 2026)

The opposite question type to prove-continuity: nothing here is done by $\varepsilon$-$\delta$. The function is handed to you as a limit, you evaluate that limit case-wise into a step function, and then you find — not verify — the discontinuity set. Expect the answer to be an infinite set.

Recognition Cues

• "$f(x)=\lim_{n\to\infty}(\ldots)$" with $n$ appearing as an exponent.
• The deliverable is “find the points of discontinuity”, i.e. a set, not a yes/no.
• The domain is an open interval — and the openness is usually load-bearing.

Solution Template

1. Name the base $t=t(x)$ and record its range. If $t$ can be $0$, say so and check the printed expression is still defined at every $n$.
2. Apply the trichotomy $t>1$, $t=1$, $0\le t<1$ and collapse $f$ to a closed form (typically a sign or step function).
3. Locate the breakpoints — where the base crosses $1$ — in terms of $x$. Substituting $u=\pi/x$ or similar is usually the cleanest route.
4. Determine the sign on each interval between breakpoints, so that $f$ is described completely.
5. Continuity off the breakpoints: $f$ is locally constant there, hence continuous — one line.
6. Discontinuity at each breakpoint: compute both one-sided limits, show they differ, and classify the jump. This is where the marks concentrate.
7. Check the endpoints of the domain. An open interval excludes points that would otherwise belong to the answer.

Worked Example

2026 Paper 2, 2026-P2-Q1c (10 marks)

Find the points of discontinuity of $f:(0,1)\to\mathbb{R}$ defined by $f(x)=\lim\limits_{n\to\infty}\dfrac{\left(1+\sin\frac{\pi}{x}\right)^n-1}{\left(1+\sin\frac{\pi}{x}\right)^n+1}$.

Source: analysis/solutions/2026-P2-Q1c.md

Step 1 — The base. Put $s=\sin\frac{\pi}{x}$ and $t=1+s$. Since $-1\le s\le1$ we have $0\le t\le2$, so $t^n\ge0$ and the denominator $t^n+1\ge1>0$: the printed quotient is well defined for every $x$ and every $n$. (Worth a line — $t$ does vanish, e.g. at $x=\tfrac23$ where $\sin\tfrac{3\pi}{2}=-1$.)

Step 2 — Trichotomy. $t>1\Rightarrow f=1$; $t=1\Rightarrow f=0$; $0\le t<1\Rightarrow f=-1$ (including $t=0$, where $t^n=0$ for all $n\ge1$). Hence

$\boxed{f(x)=\operatorname{sgn}\!\left(\sin\frac{\pi}{x}\right).}$

Step 3 — Breakpoints. With $u=\pi/x$, as $x$ runs over $(0,1)$, $u$ runs over $(\pi,\infty)$. Then $\sin u=0\iff u=k\pi$, and $u>\pi$ forces $k\ge2$, giving

$E=\left\{\tfrac1k:\ k\ge2\right\}=\left\{\tfrac12,\tfrac13,\tfrac14,\ldots\right\}.$

Note $x=1$ (which would come from $k=1$) is not in the open interval.

Step 4 — Signs. For $x\in\left(\tfrac{1}{k+1},\tfrac1k\right)$ we have $u\in(k\pi,(k+1)\pi)$, where $\sin u$ has constant sign $(-1)^k$. So $f=(-1)^k$ there, and $f=0$ on $E$.

Step 5 — Continuity off $E$. Any $x_0\notin E$ lies in one of these open intervals, on which $f$ is constant; a function constant on a neighbourhood of a point is continuous at it.

Step 6 — Discontinuity on $E$. Fix $x_0=\tfrac1k$, $k\ge2$. Then $f(x_0)=0$, while

$\lim_{x\to\frac1k^-}f=(-1)^k,\qquad \lim_{x\to\frac1k^+}f=(-1)^{k-1}=-(-1)^k,$

the right-hand interval $\left(\tfrac1k,\tfrac{1}{k-1}\right)$ being non-empty precisely because $k\ge2$. The two one-sided limits are $+1$ and $-1$ in some order, so $\lim_{x\to1/k}f$ does not exist: a first-kind (jump) discontinuity of jump $2$, with $f(1/k)=0$ sitting at the midpoint — so it is irremovable however $f(1/k)$ is redefined.

$\boxed{D=\left\{\tfrac1k:\ k=2,3,4,\ldots\right\}.}$

Two structural remarks the phrasing invites. $D$ is countably infinite and accumulates only at $x=0$, which is not in the domain; so every point of $D$ is isolated in $D$, and on any $[a,b]\subset(0,1)$ with $a>0$ there are only finitely many discontinuities (hence $f$ is Riemann integrable there). Had the domain been $[0,1)$, the point $0$ would be a far worse, second-kind discontinuity — $f$ oscillates between $\pm1$ in every neighbourhood of $0$.

Common Traps

• Doing only two branches. Omitting $0\le t<1$ (where $f=-1$) yields a different and wrong discontinuity set. All three cases must appear.
• Forgetting that the base can be $0$. The third branch must be stated as $0\le t<1$, not ”$0<t<1$, geometric” — $t=0$ genuinely occurs.
• Including $x=1$. Writing $D=\{1/k:k\ge1\}$ imports a point outside the open interval. The openness of $(0,1)$ is doing real work at both ends.
• Stopping at “discontinuous at $x=1/k$”. The question asks for the points and expects proof at each; the two one-sided limits and the classification are the marks.
• Reaching for $\varepsilon$-$\delta$. There is nothing to do with it here. The limit is evaluated case-wise; continuity off $E$ follows from local constancy in one line.

Marks-Aware Writing

At 15 marks, write a complete proof with every logical step made explicit: state $\varepsilon > 0$ given, state the goal, show the algebra, make the preliminary bound, choose $\delta$, verify the implication, and close with “since $\varepsilon$ was arbitrary.” Skipping any of these steps risks losing 3–5 marks. Roughly: setup and algebra — 4 marks; preliminary bound and $\delta$ choice — 5 marks; verification — 4 marks; conclusion — 2 marks.

Practice Set

Only one historical question on this atom (shown above).