prove limit of 2-variable using (ε,δ)-definition

Question

I have to prove using (ε,δ)-definition of limit: $$ \lim_{(x,y) \to (0,1)} ye^x = 1 $$

The problem is to work with $$|ye^x-1|$$ and show that that is less than a formula involving δ, let´s call it g(δ), formula which I will later equal to ε in order to find the δs that guarantee: $$ ||(x,y-1)||<δ \implies |ye^x-1|<ε$$

How can I apply $$ |x| \le ||(x,y-1)||$$ and $$ |y-1| \le ||(x,y-1)||$$ and not come to a dead-end in the process? I found that also useful might be $$exp |x| \ge 1 $$

Answer 1

In the following lines I elaborate 2 alternative answers based on the valuable input given by @JuliánAguirre and @SiongthyeGoh:

Based on @JuliánAguirre's answer: $$|ye^x-1|=|ye^x−e^x+e^x−1|≤e^x|y−1|+|e^x−1| $$ If $δ ≤ ln\left(1 + \frac{ϵ}{2}\right)$, then: $$ e^x|y−1|+|e^x-1|<\left(1+\frac{ϵ}{2}\right)|y−1|+\frac{ϵ}{2}<\left(1+\frac{ϵ}{2}\right)δ+\left(\frac{ϵ}{2}\right)\leϵ $$ Rearranging terms from the last inequality we get: $$δ\le\frac{\frac{ϵ}{2}}{1+\frac{ϵ}{2}} $$ Therefore the solution is: $$δ= min\left(ln\left(1 + \frac{ϵ}{2}\right);\frac{\frac{ϵ}{2}}{1+\frac{ϵ}{2}})\right) $$ Based on @SiongthyeGoh's answer: $$|ye^x-1|=|ye^x−y+y−1|≤|y||e^x−1|+|y−1| $$ If $δ ≤ \frac{ϵ}{2}$, then: $$ |y||e^x−1|+|y−1| < \left(1+\frac{ϵ}{2}\right)|e^x−1|+\frac{ϵ}{2}<\left(1+\frac{ϵ}{2}\right)|e^δ−1|+\left(\frac{ϵ}{2}\right)\leϵ $$ Rearranging terms from the last inequality we get: $$δ\le ln\left(1+\frac{\frac{ϵ}{2}}{1+\frac{ϵ}{2}}\right) $$ Therefore the solution is: $$δ= min\left(\frac{ϵ}{2};ln\left(1+\frac{\frac{ϵ}{2}}{1+\frac{ϵ}{2}}\right)\right) $$

Answer 2

Hint: $$ |y\,e^x-1|=|y\,e^x-e^x+e^x-1|\le e^x\,|y-1|+|e^x-1|. $$

Answer 3

WLOG, we can focus on $\epsilon \leq 1.$

$$|ye^x-1|=|ye^x-y+y-1| \leq |y||e^x-1|+|y-1|$$

I want to bound $|y-1|$ by $\frac{\epsilon}{2}$. So I choose to set my $\delta \leq \frac{\epsilon}{2}.$

If we look at the first term. $$|y||e^x-1|\leq (1+\frac{\epsilon}{2})|e^x-1|$$

If we purposely set $\delta \leq \ln(1+\frac{\epsilon}{3}),$

$$|y||e^x-1|\leq (1+\frac{\epsilon}{2})|e^x-1| \leq (1+\frac{\epsilon}{2})(\frac{\epsilon}{3})\leq \frac{\epsilon}{2}$$

Hence, We can choose $$\delta=\min \left(\ln\left(1+\frac{\epsilon}{3}\right),\frac{\epsilon}{2}\right).$$

Credit: Julian posted his hint first. I purposely break $|ye^x-1|$ in different form from Julian's so that you get another chance to practice bounding stuff. ;)