Worked examples · Sheet 3

Non-existence of limits via restrictions

Theory used: Restriction theorem Limit in two variables

Exercise 1

Determine whether the following limit exists

(x, y) \to (0, 0) lim \frac{x ^{2} - y ^{2}}{x ^{2} + y ^{2}} .

y = 0 : f (x, 0) = \frac{x ^{2}}{x ^{2}} = 1 x \to 0 1

x = 0 : f (0, y) = \frac{- y ^{2}}{y ^{2}} = - 1 y \to 0 - 1

1 \neq = - 1 ⟹ the limit does not exist

One computes the restrictions along the two axes. On the $x$ -axis one sets $y = 0$ : the function is constantly $1$ , so the restriction tends to $1$ . On the $y$ -axis one sets $x = 0$ : the function is constantly $- 1$ , so the restriction tends to $- 1$ . By the restriction theorem, if the limit existed every restriction would have to tend to the same value; since two restrictions produce different values, the limit does not exist.

Exercise 2

Determine whether the following limit exists

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

y = 0 : f (x, 0) = 0 x \to 0 0

y = m x : f (x, m x) = \frac{m x ^{2}}{x ^{2} + m ^{2} x ^{2}} = \frac{m}{1 + m ^{2}}

m = 1 : \frac{1}{2}, m = 0 : 0, \frac{1}{2} \neq = 0 ⟹ the limit does not exist

Along the axes the function is identically zero, because the numerator $x y$ vanishes when one of the two coordinates is zero; the axes alone are therefore not enough to reach a conclusion. One then restricts the function to the pencil of lines $y = m x$ : substituting gives $\frac{m x ^{2}}{( 1 + m ^{2} ) x ^{2}} = \frac{m}{1 + m ^{2}}$ , a value that is constant along each line but depends on the slope $m$ . For $m = 0$ the restriction tends to $0$ , while for $m = 1$ it tends to $\frac{1}{2}$ . Since two restrictions produce different limits, the limit does not exist.

Exercise 3

Determine whether the following limit exists

(x, y) \to (0, 0) lim \frac{x y ^{2}}{x ^{2} + y ^{4}} .

y = m x : f (x, m x) = \frac{m ^{2} x ^{3}}{x ^{2} + m ^{4} x ^{4}} = \frac{m ^{2} x}{1 + m ^{4} x ^{2}} x \to 0 0

x = 0 : f (0, y) = 0 y \to 0 0

x = y^{2} : f (y^{2}, y) = \frac{y ^{2} \cdot y ^{2}}{y ^{4} + y ^{4}} = \frac{1}{2}

\frac{1}{2} \neq = 0 ⟹ the limit does not exist

Restrictions at the origin: along the lines

y = m x

(dashed)

f

tends to

0

, along the parabola

x = y^{2}

it is constantly

\frac{1}{2}

Along every line $y = m x$ the restriction equals $\frac{m ^{2} x}{1 + m ^{4} x ^{2}}$ , which tends to $0$ as $x \to 0$ whatever $m$ is; the restriction to the $y$ -axis is also zero. All lines through the origin therefore produce the same value $0$ , but this does not prove the existence of the limit, because the restriction theorem provides only necessary conditions. The denominator contains $y^{4}$ , so one tries the parabola $x = y^{2}$ , which balances the degrees: substituting gives the constant value $\frac{1}{2}$ . Since the restriction to the parabola tends to $\frac{1}{2}$ and the restrictions to the lines tend to $0$ , the limit does not exist.