Epsilon delta proofs for expressions with polynomial and exponential functions

Question

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?

Answer 1

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.

Answer 2

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.