Topic 9 / 17intermediate

Limits and Continuity

Evaluate limits of functions of several variables by substitution, prove non-existence using the two-path method, and characterize continuity.

For functions of two or more variables, a limit $lim_{(x, y) \to (a, b)} f (x, y)$ requires the function value to approach the same number along **every** path leading to $(a, b)$ . This is more restrictive than a single-variable limit.

Direct substitution

f

is **continuous** at

(a, b)

— for instance, a polynomial or a rational function whose denominator does not vanish there — then

lim_{(x, y) \to (a, b)} f (x, y) = f (a, b)

. For most well-behaved problems, plug in.

Two-path method to prove non-existence

If you find two specific paths

γ_{1}, γ_{2}

approaching

(a, b)

along which

f

has different limits, the multivariable limit cannot exist. Common test paths to

(0, 0)

y = 0

x = 0

y = x

y = m x

y = x^{2}

, etc.

Strategy tips

Try linear paths first ( $y = 0$ , $y = x$ ). If they agree, try parabolic ( $y = x^{2}$ ) or polar.
If two different limits appear, you're done — limit DNE.

Continuity

f (x, y)

is **continuous at

(a, b)

** if (i)

f (a, b)

is defined, (ii) the limit exists, (iii) the two are equal. Polynomials are continuous everywhere. Rational functions are continuous wherever the denominator is nonzero. Compositions of continuous functions are continuous.

Patching for continuity

f

is given piecewise with a value

L

assigned at a point,

f

is continuous there iff the limit at that point equals

L

. To make

f

continuous, set

L =

(the limit, if it exists).

Worked examples

Example 1

Evaluate

lim_{(x, y) \to (1, 2)} (x^{2} y - 3 x y^{2} + 4)

1
Polynomial — substitute directly.
$(1)^{2} (2) - 3 (1) (2)^{2} + 4 = 2 - 12 + 4 = - 6$

Answer

- 6

Example 2

Show

lim_{(x, y) \to (0, 0)} \frac{x y}{x ^{2} + y ^{2}}

does not exist.

1
Path $y = 0$ :
$x \to 0 lim \frac{x \cdot 0}{x ^{2}} = 0$
2
Path $y = x$ :
$x \to 0 lim \frac{x \cdot x}{x ^{2} + x ^{2}} = \frac{1}{2}$
3
Two different limits → limit DNE.

Answer

Limit does not exist (paths give 0 and 1/2).

Example 3

Find

L

that makes

f

continuous at

(0, 0)

, where

f (x, y) = \frac{x ^{2} y}{x ^{2} + y ^{2}}

for

(x, y) \neq = (0, 0)

and

f (0, 0) = L

1
Bound: $∣ x^{2} y / (x^{2} + y^{2}) ∣ \leq ∣ y ∣$ since $x^{2} \leq x^{2} + y^{2}$ .
2
As $(x, y) \to (0, 0)$ , $∣ y ∣ \to 0$ , so by squeeze, $f \to 0$ .
3
Set $L = 0$ .

Answer

L = 0

Interactive visualizations

Loading visualization…

t (approach toward 0)1.00

Comparing limit values along different paths approaching

(0, 0)

y = 0: f (1.000, 0.000) = 0.000

y = x: f (1.000, 1.000) = 0.500

y = x²: f (1.000, 1.000) = 0.500

f (x, y) = x y / (x^{2} + y^{2})

approaching

(0, 0)

. Paths

y = 0

give 0;

y = x

gives 1/2 — limit DNE.

Formulas in this topic

Substitution rule

(x, y) \to (a, b) lim f (x, y) = f (a, b)

If f continuous at (a,b).

Two-path test

γ_{1} lim f \neq = γ_{2} lim f \Rightarrow lim f DNE

Disagreement along any two paths kills the limit.

Direct substitution

Two-path method to prove non-existence

Continuity

Patching for continuity

Worked examples

Interactive visualizations

Formulas in this topic

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