Proof that $\lim_{(x,y)\to (0,1)} e^{xy} = 1$

Question

Here is my proof:

Let $\epsilon > 0$ be given. There exists $\delta >0$ such that $$|e^{xy}-1| < \epsilon$$ whenever $0<|x|<\delta$ and $0<|y-1|<\delta$.

Using exponential series, we write $$|e^{xy}-1|=|(1+xy+\frac{x^2y^2}{2!}+...)-1|$$ $$=|xy+\frac{x^2y^2}{2!}+\frac{x^3y^3}{3!}+...|$$ $$\le|xy|+|\frac{x^2y^2}{2!}|+|\frac{x^3y^3}{3!}|+...$$ $$\le|xy|+|x^2y^2|+|x^3y^3|+...$$ $$\lt\delta|y|+\delta^2|y|^2+\delta^3|y|^3+...$$

Let $\delta\le\frac12\implies\frac12\lt y\lt\frac32$ $$\therefore|e^{xy}-1|\lt\frac32\delta+(\frac32\delta)^2+(\frac32\delta)^3+...$$ $$= \frac{3\delta}{2-3\delta}$$

So, we choose $$\delta = min\{\frac12,\frac{2\epsilon}{3(\epsilon+1)}\}$$

$$\implies|e^{xy}-1|\lt\epsilon$$whenever$$0<|x|<\delta $$ and $$0<|y-1|<\delta$$

Q.E.D.

I would also like to know if there are any other ways to prove this.

Answer 1

Yes, you use the composition theorem $$ f:(x,y)\to xy,\ g:z\to e^z $$ with $$ f: \mathbb{R}^2\to \mathbb{R},\ g:\mathbb{R}\to \mathbb{R} $$ For the mental exercise, it is very rewarding and I can understand it. Then I advise you to decompose this composition (i.e. general result but with $\epsilon,\delta$).
For example, first for a given $\epsilon$, find the $\delta$ corresponding to the continuity of $g$, then the $\delta_1,\delta_2$ corresponding to the continuity of $f$.

About your proof It seems clever (I only overlooked it, I must go into details), but can be made more generic by decomposing

1. For every given $\epsilon>0$, find $\delta>0$ such that $|z|<\delta\Longrightarrow |e^z-1|<\epsilon$ (you can reuse your argument proving that $|e^z-1|\leq e^{|z|}-1$)
2. For every given $\delta>0$ find $\delta_i>0,\ i=1,2$ such that $$ |x|<\delta_1\mbox{ and } |y-1|<\delta_2\Longrightarrow |xy|<\delta $$