Limits of Functions of Two Variables Examples 1

# Limits of Functions of Two Variables Examples 1

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:

## Example 1

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

Consider approaching the point $(0, 0)$ the path along the line $y = 0$. Then we get that:

(1)
\begin{align} \quad \lim_{(x, 0) \to (0,0)} \frac{x - y}{x^2 + y^2} = \lim_{x \to 0} \frac{x}{x^2} = \frac{1}{x} = \infty \end{align}

Since as $(x, y) \to (0, 0)$ along the line $y = 0$ results in $\frac{x - y}{x^2 + y^2} \to \infty$, we have that $\lim_{(x, y) \to (0, 0)}$ does not exist.

## Example 2

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

Note that for $(x,y) \neq (0,0)$ we have that:

(2)
\begin{align} 0 ≤ \frac{x^2y^2}{x^2 + y^2} = x^2 \left ( \frac{y^2}{x^2 + y^2} \right ) ≤ x^2 \end{align}

Notice that $\lim_{(x,y) \to (0, 0)} 0 = 0$ and $\lim_{(x, y) \to (0, 0)} x^2 = 0$ since both are continuous functions at $(0, 0)$. Therefore by the Squeeze Theorem we have that $\lim_{(x, y) \to (0, 0)} \frac{x^2y^2}{x^2 + y^2} = 0$.

## Example 3

Does $\lim_{(x,y) \to (1,0)} \frac{y}{x - 1}$ exist? If so, prove it and find the limit. If not, show why.

Note that if we let $w = x - 1$, then as $x \to 1$, $w \to 0$, and so we can rewrite the limit above as:

(3)
\begin{align} \quad \lim_{(x,y) \to (1,0)} \frac{y}{x - 1} = \lim_{(w,y) \to (0,0)} \frac{y}{w} \end{align}

Now notice that along the line $y = w$ we have that:

(4)
\begin{align} \quad \lim_{(w, y) \to (0, 0)} \frac{y}{y} = \lim_{(w, y) \to (0,0)} 1 = 1 \end{align}

However, along the line $y = 0$ we have that:

(5)
\begin{align} \quad \lim_{(w, 0) \to (0, 0)} \frac{0}{w} = \lim_{w \to 0} 0 = 0 \end{align}

Therefore we can approach $(0, 0)$ on two different paths that approach different values, and so $\lim_{(x,y) \to (1,0)} \frac{y}{x - 1}$ does not exist.