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Good evening,

I am working through epsilon delta proofs in my text book. I am stuck on two problems. Below are the problems and my attempt at the problems.

1) $\lim_{x\to 0}\left(xe^{2x} + 3x^{4}\right) = 0$

2) $\lim_{x\to 0}\left(xe^{-x^{2}}\right) =0$

1) So from $|f(x)-0|= |\left(xe^{2x} + 3x^{4}\right)|$, I got to $|x||e^{2x} + 3x^{3}|$ $$\text{ If }|x - 0| < 1 \text{ then } -1 < x < 1$$ and $-2 < 2x < 2$ and $-1 < x^{3} <1$ this means that: $|x||e^{2x} + 3x^{3}| < |x||e^{4} + 3|\rightarrow |x - 0|(e^{4} + 3)$ so that $\delta = \frac{\epsilon}{e^{4} + 3}$

Is this correct?

2) from $|f(x) - 0| = |x||e^{-x^{2}}|$ if $|x-0| < 1$ then $-1 < x < 1$ and $-1 < -x^{2} < 0$, thus $$|x||e^{-x^{2}}| < |x|$$ Is this approach correct?

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  • $\begingroup$ I've fixed the formatting a little bit. If you want to write text in an equation, you can use \text{}. You can also find some tips at math.meta.stackexchange.com/questions/5020/… $\endgroup$ – Arnaud D. Oct 21 '17 at 17:20
  • $\begingroup$ Thank you for that. I'm a first time user. I will get a hang of the formatting rules as I continue to post. $\endgroup$ – Destro Oct 21 '17 at 18:49
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you are almost right, when you are saying $\text{ If }|x - 0| < 1$ you are deciding upon some $\delta$ so the correct way to put the answer will be $\delta:=\min\left(1,\frac{\epsilon}{e^{4} + 3}\right)$ and likewise to the second one, but the way you found the answer is completely valid.

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  • $\begingroup$ Thank you, I omitted that by mistake. $\endgroup$ – Destro Oct 21 '17 at 18:49
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Yes, you are right. I would not say “so that $\delta=\frac\epsilon{e^4+3}$”, because it sound as if that is the only possible choice for $\delta$, I would rather write that $\delta=\frac\epsilon{e^4+3}$ will do.

In the case of the second problem, you should have added that $\delta=\epsilon$ will do.

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  • $\begingroup$ Thank you for your advise. I will make sure to do that in subsequent problems. $\endgroup$ – Destro Oct 21 '17 at 18:50
  • $\begingroup$ @Destro If my answer was useful, perhaps that you could mark it as the accepted answer. $\endgroup$ – José Carlos Santos Oct 21 '17 at 19:13

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