Limits of Functions of Two Variables Examples 2

Recall from the Limits of Functions of Two Variables page that $\lim_{(x,y) \to (a,b)} f(x,y) = L$ if:

• $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $(x, y) \in D(f)$ and $0 < \sqrt{(x-a)^2 + (y-b)^2} < \delta$ then $\mid f(x,y) - L \mid < \epsilon$.

We also noted that $\lim_{(x,y) \to (a,b)} f(x,y)$ does not exist if either:

• As $(x,y) \to (a,b)$ along some path $C$ we have that $f(x,y) \to \pm \infty$.

• For $L_1 \neq L_2$ there exists two paths $C_1$ and $C_2$ that approach $(a,b)$ such that as $(x, y) \to (a,b)$ along $C_1$ we have that $f(x,y) \to L_1$ and as $(x, y) \to (a,b)$ along $C_2$ we have that $f(x,y) \to L_2$.

To evaluate limits of two variable functions, we always want to first check whether the function is continuous at the point of interest, and if so, we can use direct substitution to find the limit. If not, then we will want to test some paths along some curves to first see if the limit does not exist. If we suspect that the limit exists after failing to show the limit does not exist, then we should attempt to utilize the definition of a limit of a two variable function and/or possibly some of the limit law theorems from the Limit Laws for Functions of Several Variables page - the Squeeze Theorem being one of the most useful.

We will now look at some more examples of evaluating two variable limits.

More examples can be found on the following pages:

• Limits of Functions of Two Variables Examples 1
• Limits of Functions of Two Variables Examples 2
• Limits of Functions of Two Variables Examples 3

Example 1

Does $\lim_{(x,y) \to (1,2)} \frac{2x - xy^2}{x + 2y}$ exist? If so, prove it and find the limit. If not, show why.

Notice that $\frac{2x - xy^2}{x + 2y}$ is only discontinuous for $x + 2y = 0$, that is $\frac{2x - xy^2}{x + 2y}$ is discontinuous on the line $y = -\frac{x}{2}$. Notice that the point $(1, 2)$ does not belong to this line, and so $\frac{2x - xy^2}{x + 2y}$ is continuous at $(1, 2)$ and:

(1)

\begin{align} \quad \lim_{(x,y) \to (1,2)} \frac{2x - xy^2}{x + 2y} = \frac{2(1) - (1)(2)^2}{(1) + 2(2)} = \frac{-2}{5} \end{align}

Example 2

Does $\lim_{(x,y) \to (0,0)} \frac{x^2y - xy}{x^2-x}$ exist? If so, prove it and find the limit. If not, show why.

We can simplify the limit above by factoring and eliminating like terms:

(2)

\begin{align} \quad \lim_{(x,y) \to (0,0)} \frac{x^2y - xy}{x^2-x} = \lim_{(x,y) \to (0,0)} \frac{(x^2 - x)y}{x^2-x} = \lim_{(x, y) \to (0, 0)} y = 0 \end{align}

So the equation $\frac{x^2y - xy}{x^2-x}$ represents the plane $z = y$ for which $x^2 - x = 0$, that is for $x \neq 0, 1$, and $\lim_{(x, y) \to (0, 0)} \frac{x^2y - xy}{x^2-x} = 0$.

Example 3

Let $f(x, y) = \left\{\begin{matrix} x + y & \mathrm{if} \: (x, y) \neq (0,0)\\ 1 & \mathrm{if} \: (x, y) = (0,0) \end{matrix}\right.$. Does $\lim_{(x,y) \to (0,0)} f(x,y)$ exist? If so, prove it and find the limit. If not, show why.

We note that $f(x, y)$ is continuous as a plane except at the point $(0, 0)$, so we suspect that the limit exists and that $\lim_{(x, y) \to (0, 0)} f(x, y) = 0$ exists. We will show this using the definition of a limit.

Let $\epsilon > 0$ be given. We want to find $\delta > 0$ such that if $(x, y) \in D(f)$ and $0 < \sqrt{x^2 + y^2} < \delta$ then $\mid f(x, y) \mid < \epsilon$. Note that:

(3)

\begin{align} \quad \mid x + y \mid ≤ \mid x \mid + \mid y \mid ≤ \sqrt{x^2} + \sqrt{y^2} ≤ \sqrt{x^2 + y^2} + \sqrt{x^2 + y^2} = 2 \sqrt{x^2 + y^2} \end{align}

Let $\delta = \frac{\epsilon}{2}$. Then for $(x, y) \in D(f)$ and $0 < \sqrt{x^2 + y^2} < \delta$ then we get that:

(4)

\begin{align} \quad \mid x + y \mid ≤ 2\sqrt{x^2 + y^2} < 2 \delta = 2 \cdot \frac{\epsilon}{2} = \epsilon \end{align}

Therefore $\lim_{(x,y) \to (0,0)} f(x,y) = 0$.