Continuity of Functions on R; Epsilon-Delta

At a Glance

• Frequency: 1 sub-part across 1 of 13 years (2018)
• 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 $x_0$.

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 $\mathbb{R}$.

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).