Limits of Functions of Two Variables Examples 4

# Limits of Functions of Two Variables Examples 4

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.

## Example 1

Evaluate the limit or prove that the limit does not exist. $\lim_{(x, y) \to (0, 0)} \frac{x^4 - 4y^2}{x^2 + 2y^2}$.

Consider this limit along the line $x = 0$. We then have that:

(1)
\begin{align} \quad \lim_{(0, y) \to (0, 0)} \frac{x^4 - 4y^2}{x^2 + 2y^2} = \lim_{y \to 0} \frac{- 4y^2}{2y^2} = \lim_{y \to 0} -2 = -2 \end{align}

Now consider this limit along the line $y = 0$. We have that:

(2)
\begin{align} \quad \lim_{(x, 0) \to (0,0)} \frac{x^4 - 4y^2}{x^2 + 2y^2} = \lim_{x \to 0} \frac{x^4}{x^2} = \lim_{x \to 0} x^2 = 0 \end{align}

Since these limits are not equal, we have that $\lim_{(x, y) \to (0, 0)} \frac{x^4 - 4y^2}{x^2 + 2y^2}$ does not exist.

## Example 2

Evaluate the limit or prove that the limit does not exist. $\lim_{(x, y) \to (0, 0)} \frac{xy}{\sqrt{x^2 + y^2}}$.

We will show that $\lim_{(x, y) \to (0, 0)} \frac{xy}{\sqrt{x^2 + y^2}} = 0$, that is show that for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $(x, y) \in D(f)$ and $0 < \sqrt{x^2 + y^2} < \delta$ then $\biggr \rvert \frac{xy}{\sqrt{x^2 + y^2}} \biggr \rvert < \epsilon$

Let $\epsilon > 0$ be given. Then:

(3)
\begin{align} \quad \biggr \rvert \frac{xy}{\sqrt{x^2 + y^2}} \biggr \rvert = \frac{\mid x \mid \sqrt{y^2}}{\sqrt{x^2 + y^2}} = \mid x \mid \sqrt{\frac{y^2}{x^2 + y^2}} ≤ \mid x \mid = \sqrt{x^2} ≤ \sqrt{x^2 + y^2} < \delta \end{align}

Therefore choose $\delta = \epsilon$. Therefore, for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $(x, y) \in D(f)$ then $\biggr \rvert \frac{xy}{\sqrt{x^2 + y^2}} \biggr \rvert < \epsilon$.

## Example 3

Evaluate the limit or prove that it does not exist. $\lim_{(x, y) \to (0, 0)} \frac{2x^2y}{x^4 + y^2}$.

Consider the limit along the line $y = x$. We have that:

(4)
\begin{align} \quad \lim_{(x, x) \to (0, 0)} \frac{2x^2y}{x^4 + y^2} = \lim_{x \to 0} \frac{2x^3}{x^4 + x^2} = 0 \end{align}

Now consider the limit along the line $y = x^2$. We have that:

(5)
\begin{align} \quad \lim_{(x, x^2) \to (0, 0)} \frac{2x^2y}{x^4 + y^2} = \lim_{x \to 0} \frac{2x^4}{x^4 + x^4} = \lim_{x \to 0} \frac{2x^4}{2x^4} = 1 \end{align}

Since these limits are not equal, we have that $\lim_{(x, y) \to (0, 0)} \frac{2x^2y}{x^4 + y^2}$ does not exist.