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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 $$exp |x| \ge 1 $$

Hanul Jeon
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Berni
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3 Answers3

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

Julián Aguirre
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  • Hi Julián, thanks for your hint. I have worked on it since and have posted an answer based on it. I would much appreciate your taking a look on it to doublecheck that it´s right. I cited you in the answer. Thank you again – Berni May 09 '16 at 21:06
  • You should complete the answer with something like: Given $\epsilon>0·$, if $\delta<\dots$ and $|x|<\delta$,$|y-1|<\delta$, then... – Julián Aguirre May 10 '16 at 08:57
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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. ;)

Siong Thye Goh
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  • thank you again for your careful explanation @SiongthyeGoh. The validity of your answer is underpinned by the initial WLOG argument, however I can´t see how you actually don´t lose generality under the assumption that ϵ is less than or equal to 1, because the delta-epsilon definition says that the δ must work for every ϵ>0. I must therefore find such δ, right? – Berni May 08 '16 at 17:26
  • That is true but we are still fine. Suppose you are given a general $\epsilon$ that is bigger than 1, use that approach that is stated to find the $\delta$ that corresponds to $\epsilon=1$. Hence you will have $|ye^x-1| \leq 1 < \epsilon$. So yup, we are usually only concern with small $\epsilon$. – Siong Thye Goh May 08 '16 at 23:06
  • Hi again @SiongthyeGoh. Would it be accurate to say that the general definition was about pointwise convergence meanwhile your solution used uniform convergence? (I got this idea reading this post). Also, I have worked on your solution during the weekend and posted a general answer that apparently works for every ϵ>0. I cited you there and I thank you again for your input. Would you mind taking a look at the answer to double check? Have a nice week – Berni May 09 '16 at 21:23
  • Yup, kinda. I use a uniform $\delta$ for $\epsilon \geq 1$. – Siong Thye Goh May 09 '16 at 21:58
  • Exactly! Learned a lot. Cya – Berni May 09 '16 at 22:06
  • remember to write the proof in forward direction too. we usually work backward to get an idea of how to choose $\delta$ (and then we hide the steps) and perform a magic show in the forward direction. – Siong Thye Goh May 09 '16 at 22:15
  • funny enough a TA recently made a similar remark when handing me back some problems regarding proofs that certain functions are bijective, and I didn't get the point of her remark; she was not as explicit. So it´s about elegance and clarity of exposition. Will keep your mathematician advice. – Berni May 09 '16 at 22:34
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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: $$δ= min\left(\frac{ϵ}{2};ln\left(1+\frac{\frac{ϵ}{2}}{1+\frac{ϵ}{2}}\right)\right) $$

Berni
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  • The second solution makes no sense (at least at the end) because $\ln(\epsilon/2)$ can be negative. – Ian May 09 '16 at 21:35
  • thank you @Ian, the natural log operator was a mistake I made when transcribing from my paper sheet to the computer. I actually used just ϵ/2 in the formula (edited now to reflect this). Do you think the worked solution is valid? Thank you! – Berni May 09 '16 at 21:59
  • Good job! It looks fine to me. – Siong Thye Goh May 09 '16 at 22:01
  • @Berni Seems fine now. – Ian May 09 '16 at 22:03