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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 $| ye^x - 1 |$.

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  • $\begingroup$ 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. $\endgroup$ – Don Thousand Apr 17 at 19:04
  • $\begingroup$ 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. $\endgroup$ – Don Thousand Apr 17 at 19:05
  • $\begingroup$ Are you allowed to use the continuity of the function $f(x) = e^x$? $\endgroup$ – avs Apr 17 at 19:08
  • $\begingroup$ 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 $n\geq 1.$ $\endgroup$ – Thomas Andrews Apr 17 at 19:19
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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 $x$.

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