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Limits & Continuity for Multivariable Functions

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# Limits & Continuity for Multivariable Functions ## 🎯 Learning Objectives After this topic you will be able to: - Evaluate limits of multivariable functions along different paths - Determine when a limit does not exist using path dependence - Use the epsilon-delta definition of continuity - Apply the squeeze theor...

Limits & Continuity for Multivariable Functions

🎯 Learning Objectives

After this topic you will be able to:
  • Evaluate limits of multivariable functions along different paths
  • Determine when a limit does not exist using path dependence
  • Use the epsilon-delta definition of continuity
  • Apply the squeeze theorem for multivariable limits
  • Understand that continuity in each variable separately does not imply joint continuity

📋 Prerequisites

  • Multivariable Functions (this week) — notation for f:RnRf: \mathbb{R}^n \to \mathbb{R}
  • Single-variable limits (BSMA1001) — limit definition, squeeze theorem
  • Limits are the foundation of calculus; continuity is essential for differentiability

1. Intuition: Approaching a Point from All Directions

In single-variable calculus, xax \to a can happen from the left or the right. For multivariable functions, xa\mathbf{x} \to \mathbf{a} can happen along infinitely many paths — straight lines, curves, spirals. For a limit to exist, the function must approach the same value no matter which path is taken. (Diagram)

2. Limits in Rn\mathbb{R}^n

2.1 Definition

Definition (Limit). limxaf(x)=L\lim_{\mathbf{x} \to \mathbf{a}} f(\mathbf{x}) = L if for every ϵ>0\epsilon > 0, there exists δ>0\delta > 0 such that:
>0<xa<δ    f(x)L<ϵ>> 0 < \|\mathbf{x} - \mathbf{a}\| < \delta \implies |f(\mathbf{x}) - L| < \epsilon >

2.2 Path Dependence

To show a limit does not exist (DNE), find two paths to a\mathbf{a} that give different limits. Common paths to test:
  • y=0y = 0 (approach along xx-axis)
  • x=0x = 0 (approach along yy-axis)
  • y=mxy = mx (approach along any line)
  • y=kx2y = kx^2 (approach along parabola)
  • x=0x = 0, y=0y = 0 (coordinate axes)
Example 1: Limit exists
f(x,y)=x2yx2+y2f(x,y) = \frac{x^2y}{x^2 + y^2}. Find lim(x,y)(0,0)f(x,y)\lim_{(x,y)\to(0,0)} f(x,y).
PathResult
y=0y = 0f(x,0)=00f(x,0) = 0 \to 0
x=0x = 0f(0,y)=00f(0,y) = 0 \to 0
y=mxy = mxf(x,mx)=x2(mx)x2+m2x2=mx1+m20f(x,mx) = \frac{x^2(mx)}{x^2+m^2x^2} = \frac{mx}{1+m^2} \to 0
y=x2y = x^2f(x,x2)=x2(x2)x2+x4=x4x2(1+x2)=x21+x20f(x,x^2) = \frac{x^2(x^2)}{x^2+x^4} = \frac{x^4}{x^2(1+x^2)} = \frac{x^2}{1+x^2} \to 0
All paths give 0. Use squeeze theorem to confirm: f(x,y)=x2yx2+y2y|f(x,y)| = \frac{x^2|y|}{x^2+y^2} \leq |y| (since x2x2+y21\frac{x^2}{x^2+y^2} \leq 1). As (x,y)(0,0)(x,y)\to(0,0), y0|y|\to0, so f0|f|\to0. Limit = 0 ✓. Example 2: Limit DNE (path dependence)
f(x,y)=xyx2+y2f(x,y) = \frac{xy}{x^2 + y^2}. Find lim(x,y)(0,0)f(x,y)\lim_{(x,y)\to(0,0)} f(x,y).
PathResult
y=0y = 0f(x,0)=00f(x,0) = 0 \to 0
x=0x = 0f(0,y)=00f(0,y) = 0 \to 0
y=mxy = mxf(x,mx)=x(mx)x2+m2x2=m1+m2f(x,mx) = \frac{x(mx)}{x^2+m^2x^2} = \frac{m}{1+m^2}
y=xy = xf(x,x)=x22x2=12f(x,x) = \frac{x^2}{2x^2} = \frac{1}{2}
y=2xy = 2xf(x,2x)=2x25x2=25f(x,2x) = \frac{2x^2}{5x^2} = \frac{2}{5}
Different lines give different limits (0, 1/2, 2/5, ...). Limit DNE.

3. Continuity

3.1 Definition

Definition (Continuity). ff is continuous at a\mathbf{a} if:
>limxaf(x)=f(a)>> \lim_{\mathbf{x} \to \mathbf{a}} f(\mathbf{x}) = f(\mathbf{a}) >

3.2 Properties

  • Sums, products, and quotients (where denominator ≠ 0) of continuous functions are continuous
  • Compositions of continuous functions are continuous
  • Polynomials in several variables are continuous everywhere
  • Rational functions are continuous where the denominator ≠ 0

4. The Squeeze Theorem

Theorem (Squeeze). If f(x)Lg(x)|f(\mathbf{x}) - L| \leq g(\mathbf{x}) near a\mathbf{a} and limxag(x)=0\lim_{\mathbf{x}\to\mathbf{a}} g(\mathbf{x}) = 0, then limxaf(x)=L\lim_{\mathbf{x}\to\mathbf{a}} f(\mathbf{x}) = L. Example 3: Using squeeze theorem
f(x,y)=x3x2+y2f(x,y) = \frac{x^3}{x^2 + y^2}. Show lim(x,y)(0,0)f(x,y)=0\lim_{(x,y)\to(0,0)} f(x,y) = 0.
f(x,y)=x3x2+y2=xx2x2+y2x|f(x,y)| = \frac{|x|^3}{x^2+y^2} = |x| \cdot \frac{x^2}{x^2+y^2} \leq |x|.
Since x0|x| \to 0 as (x,y)(0,0)(x,y)\to(0,0), by squeeze theorem f0|f| \to 0. ✓

5. Common Pitfalls

❌ Pitfall 1: Checking only lines

Showing the limit along every line agrees does not prove the limit exists — there could be a parabolic or spiral path with a different limit.

❌ Pitfall 2: Using polar coordinates incorrectly

Converting to polar can help, but you must show the limit is independent of θ\theta as r0r\to0.

❌ Pitfall 3: Separate continuity ≠ joint continuity

f(x,y)=xyx2+y2f(x,y) = \frac{xy}{x^2+y^2} (with f(0,0)=0f(0,0)=0) is continuous in xx for each yy and in yy for each xx, but not continuous in (x,y)(x,y) jointly.

6. Formula Summary Table

ConceptKey Idea
Multivariable limitlimxaf(x)=L\lim_{\mathbf{x}\to\mathbf{a}} f(\mathbf{x}) = L if all paths agree
Path testIf two paths give different limits, limit DNE
Continuitylimf(x)=f(a)\lim f(\mathbf{x}) = f(\mathbf{a})
Squeeze theorem$

7. 📝 Practice Questions

Q1: Limit along lines
f(x,y)=x2y2x2+y2f(x,y) = \frac{x^2 - y^2}{x^2 + y^2}. Does lim(x,y)(0,0)f\lim_{(x,y)\to(0,0)} f exist?
Solution: Along y=0y=0: f(x,0)=11f(x,0) = 1 \to 1. Along x=0x=0: f(0,y)=11f(0,y) = -1 \to -1. Different limits ⇒ DNE. Q2: Limit exists
f(x,y)=sin(xy)yf(x,y) = \frac{\sin(xy)}{y}. Does lim(x,y)(0,0)f\lim_{(x,y)\to(0,0)} f exist?
Solution: Along y=0y=0? Not defined. Use the fact sin(u)/u1\sin(u)/u \to 1. Write sin(xy)y=xsin(xy)xy\frac{\sin(xy)}{y} = x \cdot \frac{\sin(xy)}{xy}. As (x,y)(0,0)(x,y)\to(0,0), xy0xy \to 0, so sin(xy)xy1\frac{\sin(xy)}{xy} \to 1. Therefore f(x,y)01=0f(x,y) \to 0 \cdot 1 = 0. Limit = 0. Q3: Limit DNE — parabolic path
f(x,y)=x2yx4+y2f(x,y) = \frac{x^2y}{x^4 + y^2}. Show lim(x,y)(0,0)f\lim_{(x,y)\to(0,0)} f DNE.
Solution: Along y=mxy = mx: f(x,mx)=mx3x4+m2x2=mxx2+m20f(x,mx) = \frac{mx^3}{x^4 + m^2x^2} = \frac{mx}{x^2 + m^2} \to 0 for m0m \neq 0. Along y=x2y = x^2: f(x,x2)=x2(x2)x4+x4=x42x4=12f(x,x^2) = \frac{x^2(x^2)}{x^4 + x^4} = \frac{x^4}{2x^4} = \frac{1}{2}. Different limits (0 vs 1/2) ⇒ DNE. Q4: Squeeze theorem
f(x,y)=x2y2x2+y2f(x,y) = \frac{x^2y^2}{x^2 + y^2}. Show limf=0\lim f = 0 at (0,0)(0,0).
Solution: 0x2y2x2+y2(x2+y2)(x2+y2)x2+y2=x2+y20 \leq \frac{x^2y^2}{x^2+y^2} \leq \frac{(x^2+y^2)(x^2+y^2)}{x^2+y^2} = x^2+y^2 (since x2y2(x2+y2)2/4x^2y^2 \leq (x^2+y^2)^2/4, actually x2y2(x2+y2)2/4x^2y^2 \leq (x^2+y^2)^2/4). Better: 0x2y2x2+y2x2(x2+y2)x2+y2=x20 \leq \frac{x^2y^2}{x^2+y^2} \leq \frac{x^2(x^2+y^2)}{x^2+y^2} = x^2 (since y2x2+y2y^2 \leq x^2+y^2). As x0x\to0, x20x^2\to0, so limit = 0 by squeeze. Q5: Continuity
Is
>f(x,y)={xyx2+y2(x,y)(0,0)0(0,0)>> f(x,y) = \begin{cases} \frac{xy}{x^2+y^2} & (x,y)\neq(0,0) \\ 0 & (0,0) \end{cases} >
continuous at (0,0)(0,0)?
Solution: Earlier we showed the limit DNE along different lines. So limff(0,0)=0\lim f \neq f(0,0)=0. ff is not continuous at (0,0)(0,0). Q6: Making a function continuous
Define f(0,0)f(0,0) to make f(x,y)=x2yx2+y2f(x,y) = \frac{x^2y}{x^2+y^2} continuous.
Solution: From Example 1, lim(x,y)(0,0)f=0\lim_{(x,y)\to(0,0)} f = 0. Define f(0,0)=0f(0,0) = 0. Q7: Polar coordinates
Use polar coordinates to evaluate lim(x,y)(0,0)x3x2+y2\lim_{(x,y)\to(0,0)} \frac{x^3}{x^2+y^2}.
Solution: x=rcosθx = r\cos\theta, y=rsinθy = r\sin\theta. x3x2+y2=r3cos3θr2=rcos3θ\frac{x^3}{x^2+y^2} = \frac{r^3\cos^3\theta}{r^2} = r\cos^3\theta. As r0r \to 0, rcos3θ0r\cos^3\theta \to 0 regardless of θ\theta. Limit = 0. Q8: Continuity of rational function
Where is f(x,y)=x+yxyf(x,y) = \frac{x+y}{x-y} continuous?
Solution: ff is a rational function, continuous everywhere except where denominator =0=0, i.e., xy=0x-y=0 or y=xy=x. It's continuous on R2{y=x}\mathbb{R}^2 \setminus \{y=x\}. Q9: Limit of composition
lim(x,y)(0,0)sin(x2+y2x2+2y2)\lim_{(x,y)\to(0,0)} \sin\left(\frac{x^2+y^2}{x^2+2y^2}\right)
Solution: x2+y2x2+2y2\frac{x^2+y^2}{x^2+2y^2} is bounded between 12\frac{1}{2} and 11 (when y=0y=0, it's 11; when x=0x=0, it's 1/21/2). But the limit as (x,y)(0,0)(x,y)\to(0,0) depends on path. Along y=0y=0: ratio =1=1, sin(1)\sin(1). Along x=0x=0: ratio =1/2=1/2, sin(1/2)\sin(1/2). Different limits ⇒ limit of the composition DNE. Q10: Epsilon-delta proof
Prove lim(x,y)(0,0)3x2y+1=1\lim_{(x,y)\to(0,0)} 3x - 2y + 1 = 1 using ϵ\epsilon-δ\delta.
Proof: Given ϵ>0\epsilon > 0, choose δ=ϵ/5\delta = \epsilon/5 (or ϵ/3\epsilon/3). If (x,y)=x2+y2<δ\|(x,y)\| = \sqrt{x^2+y^2} < \delta, then 3x2y+11=3x2y3x+2y5x2+y2<5δ=ϵ|3x-2y+1-1| = |3x-2y| \leq 3|x| + 2|y| \leq 5\sqrt{x^2+y^2} < 5\delta = \epsilon. Thus limf=1\lim f = 1. ✓

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