Determine continuity at $(0,0)$ of the function $\frac{e^{xy}-1}{|x|+|y|}$.

Question

Prepping for a calculus exam. In a past exam I saw the following question:

Determine whether the following function is continuous at $(0,0)$: $$ f(x,y)= \begin{cases} \frac{e^{xy}-1}{|x|+|y|}, & \text{if $(x,y)\neq$(0,0)} \\ 0 & \text{if $(x,y)=$(0,0)} \end{cases} \ $$

Not sure where to begin.

Answer 1

1. I will show that $g(x,y)=\frac{xy}{|x|+|y|}$ is continuous at the origin (putting $g(0,0)=0$, of course). If we write it in polar coordinates $x=r\cos\theta,y=r\sin\theta$, we get: $$ g(r,\theta)=\frac{r\cos\theta\sin\theta}{|\cos\theta|+|\sin\theta|}. $$ Note that the denominator is nonzero, i.e. $|\cos\theta|+|\sin\theta|\ge C$ for some $C>0$ (exercise). Therefore, $$ |g-0|\le \frac{r|\cos\theta\sin\theta|}{C}\le \frac{r}{2C}=\frac{1}{2C}\sqrt{x^2+y^2}. $$ We can then use the $\epsilon/\delta$ definition of continuity (exercise).

2. Here's two approaches you can try for $f(x,y)$:

a. $e^{xy}=1+xy+\frac{(xy)^2}{2!}+\dots$ (Taylor Series).

b. $e^{xy}-1=xy\int_0^1e^{(xy)s}ds$ (FTC).

Answer 2

In 2D you can approach the origin from more directions than just from the left or right in 1D.

You need to show that any such approach gives a limit value which is the same as the function value. Or find a counter example.

Answer 3

Hint:

For $y \neq 0$

$$|\frac{e^{xy}-1}{|x|+|y|} | \leq |\frac{e^{xy}-1}{|y|}|=| \frac{e^{xy}-1}{y}|$$

Since for positive numbers $a$ and $b$, a smaller denominator $b$ of $\frac{a}{b}$ means a larger result, keeping $a$ fixed but allowing $b$ to vary.

$$\lim_{(x,y) \to (0,0)} \frac{e^{xy}-1}{y}=\lim_{(x,y) \to (0,0)} x \frac{e^{xy}-1}{xy}=(0)(1)=0$$ By squeeze theorem the limit must be zero (work out $y=0$, $x \neq 0$ separately and also $y \neq 0$, $x=0$).