Limits of Functions of Two Variables Examples 3

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

Does $\lim_{(x,y) \to (0,0)} \frac{y^3}{x^2+y^2}$ exist? If so, prove it and find the limit. If not, show why.

Consider approaching $(0, 0)$ along the lines $y = mx$ for $m \in \mathbb{R}$. Then we get that:

(1)

\begin{align} \quad \lim_{(x, mx) \to (0, 0)} \frac{m^3x^3}{x^2 + m^2x^2} = \lim_{x \to 0} \frac{m^3x^3}{(m^2 + 1)x^2} = \lim_{x \to 0} \frac{m^3x}{m^2 + 1} = 0 \end{align}

Therefore approaching $(0, 0)$ along any line in the form $y = mx$ approaches $0$. Now let's see what $\frac{y^3}{x^2+y^2}$ approaches as $(x, y) \to (0, 0)$ with parabolas in the form $y = mx^2$

(2)

\begin{align} \quad \lim_{(x, y) \to (0, 0)} \frac{m^3 x^6}{x^2 + m^2 x^4} = \lim_{x \to 0} \frac{m^3 x^6}{x^2 + m^2 x^4} = \lim_{x \to 0} \frac{m^3 x^4}{1 + m^2x^2} = 0 \end{align}

So we haven't proven that $\lim_{(x,y) \to (0,0)} \frac{y^3}{x^2+y^2}$ does not exist. Let's try seeing what happens as we approach $(0, 0)$ along the cubics $y = mx^3$. Then we get that:

(3)

\begin{align} \quad \lim_{(x, y) \to (0, 0)} \frac{m^3x^9}{x^2 + m^2x^6} = \lim_{x \to 0} \frac{m^3x^7}{1 + m^2x^4} = 0 \end{align}

So we suspect now that $\lim_{(x,y) \to (0,0)} \frac{y^3}{x^2+y^2} = 0$. Let's try to prove this using the definition of a limit. We want to show that $\forall \epsilon > 0$ $\exists \delta > 0$ such that if $(x, y) \in D(f)$ and $0 < \sqrt{x^2 + y^2} < \delta$ then $\biggr \rvert \frac{y^3}{x^2+y^2} \biggr \rvert < \epsilon$. Note that:

(4)

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

So let $\delta = \epsilon$. Then if $(x, y) \in D(f)$ and $0 < \sqrt{x^2 + y^2} < \delta = \epsilon$ then we have that:

(5)

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

Therefore $\lim_{(x,y) \to (0,0)} \frac{y^3}{x^2+y^2} = 0$.