Continuity of real functions

At a Glance

• Frequency: 2 sub-parts across 2 of 14 years (2019, 2024)
• Priority tier: T3
• Marks (count): 10 (2)
• Average solve time: ~6 min
• Difficulty mix: easy 2
• Section: A | Dominant type: computation

Why This Chapter Matters

Continuity questions appear in Section A (compulsory) and are always 10 marks — meaning a clean, structured answer wins full marks in five minutes and frees time for harder problems. The two question types are narrow: either discuss one-sided limits to classify a discontinuity, or evaluate a $0/0$ limit to assign a value that makes a function continuous. Both types reward a fixed three-step scaffold, and both appeared in recent papers (2019, 2024), making this a reliable source of easy marks.

Minimum Theory

Definition of continuity. A function $f$ is continuous at $x=a$ if: (i) $f(a)$ is defined, (ii) $\lim_{x\to a}f(x)$ exists, and (iii) $\lim_{x\to a}f(x)=f(a)$. Equivalently, $\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=f(a)$.

Types of discontinuity. If both one-sided limits exist but are unequal, $f$ has a jump discontinuity. If both one-sided limits equal a finite value $L\ne f(a)$ (or $f(a)$ is undefined), the discontinuity is removable and $f$ can be redefined as $f(a)=L$ to make it continuous.

Standard limits. $\displaystyle\lim_{h\to 0}\frac{\sin h}{h}=1$. For $0/0$ limits, use factoring, substitution $x=a+h$, or L’Hôpital’s rule. For functions involving $e^{1/x}$: as $x\to 0^+$, $1/x\to+\infty$ and $e^{1/x}\to+\infty$; as $x\to 0^-$, $1/x\to-\infty$ and $e^{1/x}\to 0$. The behaviour is asymmetric — always treat the two sides separately.

Question Archetypes

continuity-discussion (1 question(s); 2024)

Recognition Cues

• “Discuss continuity … for all values of $x$” with a piecewise definition.
• The function involves $e^{-1/x}$ or $e^{1/x}$: the key is that $1/x$ changes sign as $x$ passes through $0$.
• A specific value is assigned at the boundary point (e.g.\ $f(0)=0$).

Solution Template

1. Away from the special point. Show $f$ is continuous for $x\ne 0$ (usually: composition of continuous functions, denominator nonzero).
2. Left-hand limit at $x=0$. Determine the sign of the exponent as $x\to 0^-$ and evaluate $e^{\pm\infty}$.
3. Right-hand limit at $x=0$. Same analysis for $x\to 0^+$.
4. Compare limits with $f(0)$. State whether left-/right-continuous, and classify the discontinuity (jump, removable, or continuity).

Worked Example

2024 Paper 1, 2024-P1-Q1c (10 marks)

Discuss the continuity of $f(x)=\dfrac{1}{1-e^{-1/x}}$ for $x\ne 0$, $f(0)=0$, for all values of $x$.

Step 1 — Away from $x=0$. For $x\ne 0$, $-1/x$ is finite and nonzero, so $e^{-1/x}\ne 1$, and the denominator is nonzero. Hence $f$ is a composition of continuous functions and is continuous at every $x\ne 0$.

Step 2 — Left-hand limit. As $x\to 0^-$, $-1/x\to+\infty$, so $e^{-1/x}\to+\infty$, and

$\lim_{x\to 0^-}f(x)=\frac{1}{1-\infty}=0.$

Step 3 — Right-hand limit. As $x\to 0^+$, $-1/x\to-\infty$, so $e^{-1/x}\to 0$, and

$\lim_{x\to 0^+}f(x)=\frac{1}{1-0}=1.$

Step 4 — Compare with $f(0)=0$.

Left limit $=0=f(0)$: $f$ is left-continuous at $0$.

Right limit $=1\ne f(0)=0$: $f$ is not right-continuous at $0$.

$\boxed{\;f\text{ is continuous for all }x\ne 0;\;f\text{ has a jump discontinuity at }x=0\text{ (jump }=1\text{).}\;}$

Common Traps

• Computing only $\lim_{x\to 0}f(x)$ (without distinguishing left and right) loses all marks — the whole difficulty lies in the asymmetry.
• As $x\to 0^-$: $-1/x$ is positive (since $x<0$), giving $e^{-1/x}\to+\infty$, not zero. Getting the sign wrong reverses the answer.
• The denominator $1-e^{-1/x}\to 1-\infty=-\infty$, so $f\to 1/(-\infty)=0$. Don’t write $1/0$.
• Always state the type of discontinuity (jump, with the jump size) for full marks.

removable-limit (1 question(s); 2019)

Recognition Cues

• “Let $f$ be continuous on $[\,\cdot\,,\cdot\,]$; find $f(a)$” where the given formula is $0/0$ at $x=a$.
• The continuity hypothesis forces $f(a)=\lim_{x\to a}f(x)$: the problem reduces entirely to evaluating the limit.
• Typical structure: numerator has a double zero at $a$ (e.g.\ $\cos^2 x$ near $x=\pi/2$) while denominator has a simple zero — so the limit is $0$.

Solution Template

1. Invoke continuity. State: since $f$ is continuous at $x=a$, $f(a)=\lim_{x\to a}f(x)$.
2. Identify $0/0$. Show numerator and denominator both vanish at $x=a$.
3. Substitute $x=a+h$, $h\to 0$. Rewrite numerator and denominator in terms of $h$.
4. Evaluate the limit. Factor, cancel, and use standard limits (e.g.\ $\sin h/h\to 1$).
5. State the value and box the answer.

Worked Example

2019 Paper 1, 2019-P1-Q1a (10 marks)

Let $f:\left[0,\dfrac{\pi}{2}\right]\to\mathbb R$ be continuous and $f(x)=\dfrac{\cos^2 x}{4x^2-\pi^2}$ for $0\le x<\dfrac{\pi}{2}$. Find $f\!\left(\dfrac{\pi}{2}\right)$.

Step 1 — Invoke continuity. Since $f$ is continuous on $\left[0,\frac{\pi}{2}\right]$,

$f\!\left(\tfrac{\pi}{2}\right)=\lim_{x\to\frac{\pi}{2}^-}\frac{\cos^2 x}{4x^2-\pi^2}.$

Step 2 — Identify $0/0$. As $x\to\frac{\pi}{2}$: $\cos x\to 0$ so $\cos^2 x\to 0$; $4x^2-\pi^2=(2x-\pi)(2x+\pi)\to 0\cdot(2\pi)=0$.

Step 3 — Substitute $x=\frac{\pi}{2}+h$, $h\to 0$. Then $\cos x=\cos\!\left(\frac{\pi}{2}+h\right)=-\sin h$, so $\cos^2 x=\sin^2 h$. Also

$4x^2-\pi^2=(2x-\pi)(2x+\pi)=2h\cdot(2\pi+2h)=4h(\pi+h).$

Hence

$\frac{\cos^2 x}{4x^2-\pi^2}=\frac{\sin^2 h}{4h(\pi+h)}=\frac{1}{4(\pi+h)}\cdot\left(\frac{\sin h}{h}\right)^{\!2}\cdot h.$

Step 4 — Evaluate. Using $\sin h/h\to 1$ as $h\to 0$:

$\lim_{h\to 0}\frac{1}{4(\pi+h)}\cdot 1\cdot h=\frac{1}{4\pi}\cdot 0=0.$

$\boxed{\;f\!\left(\tfrac{\pi}{2}\right)=0.\;}$

Common Traps

• $f(\pi/2)$ is defined by the limit (continuity), not by the formula — the formula is undefined at $\pi/2$.
• $\cos^2 x$ vanishes to second order at $\pi/2$ while the denominator vanishes to first order (one factor $(2x-\pi)$): the limit is $0$, not $1/(4\pi)$.
• Factor the denominator as $(2x-\pi)(2x+\pi)$; expanding it makes the zero structure harder to see.
• L’Hôpital’s rule works as a check: $\frac{-\sin 2x}{8x}\to\frac{-\sin\pi}{4\pi}=0$ ✓.

Marks-Aware Writing

Both exam questions are 10 marks. For the continuity-discussion type: the four steps (continuity away from special point, left limit, right limit, comparison with $f(0)$) together account for all marks — omitting the step that shows continuity away from zero costs 2 marks, and treating both one-sided limits identically costs 5 marks. Always name the discontinuity type. For the removable-limit type: stating “by continuity, $f(\pi/2)=\lim\cdots$” is worth 2 marks; the substitution and limit evaluation account for the remaining 8. A student who writes $1/(4\pi)$ (first-order/first-order confusion) earns at most 3 marks.

Practice Set

• 2019-P1-Q1b (10 m) — — Hint: differentiability implies continuity but not vice versa; construct or identify the point where continuity fails.
• 2019-P1-Q2a (15 m) — — Hint: apply the intermediate value theorem or squeeze theorem after establishing the limit.