# Finding upper bounds for $e^x$ in limit exercises

I'm trying to prove by definition

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

How can I find upper bounds for $$| ye^x - 1 |$$? I know that $$| ye^x - 1 | = | y(e^x - 1) + (y - 1) |$$ But I never know how to get rid of things that involve $$e^x$$ such as $$| ye^x - 1 |$$.

• Don't re-ask a question just because you haven't received a full answer. I didn't give you a full answer because I wanted you to think about that expansion. Not because I wanted you to ask the question again. – Don Thousand Apr 17 at 19:04
• Look at the approach provided by the answerer of your other problem. I'm assuming these are pset problems. Try to solve them and learn from them. – Don Thousand Apr 17 at 19:05
• Are you allowed to use the continuity of the function $f(x) = e^x$? – avs Apr 17 at 19:08
• For $|x|<1$ you can show that $|e^x-1|\leq |x|(e-1),$ which you can see from the power series of $e^x$ and that $|x^n|\leq |x|$ for all $n\geq 1.$ – Thomas Andrews Apr 17 at 19:19

## 1 Answer

Write $$|y (e^x - 1)| = |y| |e^x - 1|.$$ Since $$y$$ tends to $$1$$, we know that, eventually, $$|y| < 2$$. Therefore, $$|y| |e^x - 1| < 2 |e^x - 1|.$$ Now work with $$|e^x - 1|$$. It helps to use the fact that the function $$f(x) = e^x - 1$$ is continuous at all $$x$$.