Functions of two/three variables: limits, continuity

At a Glance

• Frequency: 4 sub-parts across 3 of 13 years (2015, 2016, 2019)
• Priority tier: T3
• Marks (count): 10 (3), 12 (1)
• Average solve time: ~7 min
• Difficulty mix: medium 2, easy 1, hard 1
• Section: A | Dominant type: proof

Why This Chapter Matters

This atom covers the three fundamental tools for analysing multi-variable functions at a point: the path test (to prove a limit does not exist), the $\varepsilon$-$\delta$ argument (to prove it does), and the factorisation trick (to handle removable discontinuities). Each has appeared as a 10–12 mark Section A question. The path-test and epsilon-delta problems are particularly formulaic — once you know the recipe, they take under 8 minutes.

Minimum Theory

Limit in $\mathbb{R}^2$. $\lim_{(x,y)\to(a,b)} f(x,y) = L$ means: for every $\varepsilon > 0$ there is $\delta > 0$ such that $0 < \sqrt{(x-a)^2+(y-b)^2} < \delta \Rightarrow |f(x,y)-L| < \varepsilon$.

Path test (prove non-existence). If $f$ has different limits along two different paths through $(a,b)$, then $\lim_{(x,y)\to(a,b)} f$ does not exist. Useful paths: lines $y = kx$; parabolas $y = tx^2$; along axes ($x=0$ or $y=0$).

Epsilon-delta argument (prove existence). Convert to polar: $x = r\cos\theta$, $y = r\sin\theta$. If $|f(r,\theta)| \le M \cdot r$ for some constant $M$ uniform in $\theta$, then choosing $\delta = \varepsilon/M$ proves $f \to 0$.

Continuity. $f$ is continuous at $(a,b)$ if $\lim_{(x,y)\to(a,b)} f(x,y) = f(a,b)$.
Implication: continuity of $f$ at $(a,b)$ implies continuity of the section functions $f(x,b)$ and $f(a,y)$ at $x=a$ and $y=b$ (the converse is false).

Question Archetypes

path-test-discontinuity (1 question; 2015)

Recognition Cues

• Piecewise function, $f = [\text{expression}]$ for $(x,y) \ne (0,0)$ and $f = 0$ at origin.
• “Examine continuity and differentiability.”
• The expression has $(x^2+y^2)$ or $(x-y)$ in the denominator.

Solution Template

1. Try the $x$-axis path ($y = 0$): compute $\lim_{x\to0} f(x,0)$.
2. Try another path ($y = kx$ or $y = tx^2$): compute the limit. If different from step 1, limit does not exist, $f$ is discontinuous.
3. Conclude non-differentiability (discontinuity $\Rightarrow$ not differentiable).
4. Check partial derivatives at $(0,0)$ by definition (they may or may not exist separately).

Worked Example

2015 Paper 1, 2015-P1-Q4d (12 marks)

Examine continuity and differentiability of $f(x,y) = (x^2-x\sqrt{y})/(x^2+y)$ at $(0,0)$, where $f(0,0)=0$.

Path $y = 0$ ($x \to 0^+$): $f(x,0) = x^2/x^2 = 1$.

Path $y = tx^2$ ($x \to 0^+$, $t > 0$): $f(x,tx^2) = (x^2 - x\cdot x\sqrt{t})/(x^2+tx^2) = (1-\sqrt{t})/(1+t)$.

This depends on $t$ (e.g. $t=0 \to 1$; $t=1 \to 0$; $t=4 \to -1/5$). The limit is path-dependent.

Conclusion: $f$ is not continuous at $(0,0)$ (limit does not equal $f(0,0) = 0$, and doesn’t even exist). Hence $f$ is not differentiable at $(0,0)$.

Partials: $f_y(0,0) = 0$ (from $f(0,k) = 0$), but $f_x(0,0)$ does not exist (from $f(h,0) = 1$ for $h \ne 0$).

Common Traps

• The natural path test is parabolic $y = tx^2$, not linear $y = kx$. Linear paths sometimes give the same limit (false positive); parabolic paths expose the true direction-dependence.
• Discontinuity at a point does not stop the partial derivatives existing: $f_y(0,0) = 0$ here even though $f$ is discontinuous. The failure of $f_x(0,0)$ to exist is an additional irregularity.

epsilon-delta-continuity (1 question; 2016)

Recognition Cues

• “Find $\delta > 0$ such that $|f(x,y) - f(0,0)| < 0.01$ whenever $\sqrt{x^2+y^2} < \delta$.”
• The function is given with a piecewise definition at $(0,0)$.
• Requires an explicit $\delta$, not just an existence argument.

Solution Template

1. Switch to polar: $x = r\cos\theta$, $y = r\sin\theta$, so $|f| = r\cdot|g(\theta)|$ for some bounded $g$.
2. Bound $|g(\theta)|$ uniformly: find $M = \max_\theta |g(\theta)|$.
3. Choose $\delta$: if $|f| \le Mr$, take $\delta = \varepsilon/M$ (here $\delta = 0.01/M$).

Worked Example

2016 Paper 1, 2016-P1-Q3b (15 marks)

Find $\delta$ for $|f(x,y)| < 0.01$ where $f = (2x^4y - 5x^2y^3 + y^5)/(x^2+y^2)^2$, $f(0,0)=0$.

(The numerator has degree 5, denominator degree 4 — this is the corrected version with $5x^2y^3$.)

Polar: numerator $= r^5(2\cos^4\theta\sin\theta - 5\cos^2\theta\sin^3\theta + \sin^5\theta)$, denominator $= r^4$. So $|f| = r\cdot|g(\theta)|$ where $|g(\theta)| \le 1$ (check: $g(\theta) = \sin\theta(8\sin^4\theta-9\sin^2\theta+2)$, max = 1 at $\sin\theta = \pm1$).

So $|f| \le r$. Choose $\delta = 0.01$.

$\boxed{\delta = 0.01.}$

(A conservative elementary bound $|f| \le 8r$ gives $\delta = 1/800$; both are correct.)

Common Traps

• The bound $|f| \le Mr$ must hold uniformly in $\theta$ — not just for specific paths.
• The 2016 question as printed has a typo ($5x^2y^2$ instead of $5x^2y^3$); the non-homogeneous version is discontinuous at the origin and no $\delta$ exists. Solve the homogeneous version.

slice-continuity-proof (1 question; 2019)

Recognition Cues

• “If $f(x,y)$ is continuous at $(a,b)$, show the functions $f(x,b)$ and $f(a,y)$ are continuous at $a$ and $b$ respectively.”
• Pure $\varepsilon$-$\delta$ proof; no computation.

Worked Example

2019 Paper 1, 2019-P1-Q1b (10 marks)

Show: $f$ continuous at $(a,b)$ $\Rightarrow$ $f(x,b)$ continuous at $a$ and $f(a,y)$ continuous at $b$.

Proof. Let $g(x) = f(x,b)$. For any $\varepsilon > 0$, continuity of $f$ gives $\delta > 0$ with $\sqrt{(x-a)^2+(y-b)^2} < \delta \Rightarrow |f(x,y)-f(a,b)| < \varepsilon$.

For any $x$ with $|x-a| < \delta$: the point $(x,b)$ satisfies $\sqrt{(x-a)^2+0^2} = |x-a| < \delta$, so $|g(x)-g(a)| = |f(x,b)-f(a,b)| < \varepsilon$. Hence $g$ is continuous at $a$. The argument for $h(y) = f(a,y)$ is identical. $\square$

Remark. The converse is false: $f(x,y) = xy/(x^2+y^2)$ (with $f(0,0)=0$) has $f(x,0) \equiv 0$ and $f(0,y) \equiv 0$ (both continuous), but $f(x,x) = 1/2 \not\to 0$ — jointly discontinuous.

Common Traps

• The key computation is $\|(x,b)-(a,b)\| = |x-a|$ (the second coordinate is fixed). This is the only non-trivial step.
• The converse must not be proved — mention the counterexample to show awareness.

removable-via-factoring (1 question; 2019)

Recognition Cues

• Function has denominator $x-y$ (or similar) and the question asks to show it is continuous and differentiable at a specific point with $x \ne y$.
• Factoring the numerator completely removes the singularity at that point.

Worked Example

2019 Paper 2, 2019-P2-Q1b (10 marks)

Show $f(x,y) = (x^2-y^2)/(x-y)$ for $(x,y) \ne (1,\pm 1)$, $f=0$ at $(1,\pm1)$, is continuous and differentiable at $(1,-1)$.

Factoring: $(x^2-y^2)/(x-y) = x+y$ for $x \ne y$. Since $1 \ne -1$, on a neighbourhood of $(1,-1)$ the denominator is nonzero and $f(x,y) = x+y$.

Continuity: $\lim_{(x,y)\to(1,-1)}(x+y) = 1+(-1) = 0 = f(1,-1)$. Continuous.

Differentiability: $f(1+h_1,-1+h_2) - f(1,-1) = h_1+h_2$ exactly. The remainder with gradient $(1,1)$ is 0 = $o(\|h\|)$. Differentiable with $\nabla f = (1,1)$.

Common Traps

• Verify $x \ne y$ at the target point before factoring. Here $1 \ne -1$ ✓.
• The function also has a special value at $(1,1)$ on the diagonal, but that point is irrelevant to behaviour at $(1,-1)$.

Marks-Aware Writing

10-mark proof (slice continuity): Four lines: state the $\delta$ from joint continuity, observe $\|(x,b)-(a,b)\|=|x-a|$, apply the joint $\delta$, conclude. Then one sentence on the counterexample to the converse.

12-mark path test + partial derivatives: Show two paths giving different limits (2 lines each). State discontinuity. State non-differentiability follows. Then compute $f_x(0,0)$ and $f_y(0,0)$ by definition. Full-marks answer shows all four items.

Practice Set

• 2013-P1-Q3b (15 m) — mixed partials Schwarz failure;
• 2019-P1-Q4c-i (12 m) — continuity analysis;
• 2018-P1-Q3b (12 m) —
• 2017-P1-Q3c (15 m) — mixed partials at origin;