$\lim_{x\rightarrow\infty} e^{-x^2}e^{2x}$ (epsilon-delta proof)?

Question

I am struggling to develop an epsilon-delta proof for the following:

$\lim_{x\rightarrow\infty} e^{-x^2} e^{2x} = L$ (unknown, believed to be 0)

I am aware that to do so, we must show $\exists\beta \ s.t. \beta < x \implies |e^{-x^2} e^{2x} - L| < \epsilon $

EDIT: I believe the following may be appropriate. Let L = 0 (as per your comments/suggestions).

\begin{align*} |e^{-x^2} e^{2x}| &< \epsilon \\ e^{-x^2} e^{2x} &< \epsilon \\ \text{ln}(e^{-x^2 + 2x}) &< \text{ln}(\epsilon) \\ -x^2+2x &< \text{ln}(\epsilon) \\ \text{ln}\bigg(\frac{1}{\epsilon}\bigg) &< x^2 - 2x < x^2 \\ \sqrt{\text{ln}\bigg(\frac{1}{\epsilon}\bigg)} &< x \end{align*}

EDIT (Last) I just realized the preceding is valid only $\forall x > 0 \ , \ x \in \mathbb{R}$ as $x^2 -2x < x^2$ only for the aforementioned domain. Technically speaking, does this make the proof incorrect?

Answer 1

After you guess the limit is zero, you should prove it, so with your comment $$\frac{1}{e^{x(x-2)}}<\epsilon$$ and $(x-1)^2-1>\ln\dfrac{1}{\epsilon}$ and with $\epsilon<e$ then $x>\sqrt{\ln\dfrac{1}{\epsilon}+1}+1$

Answer 2

$e^{-x^2}e^{2x}=e^{-x^2+2x}$, $lim_{\infty}-x^2+2x=-\infty$ implies that $lim_{\infty}e^{-x^2+2x}=0$.

Answer 3

$0\le e^{-x^2}e^{2x}=e^{-x^2+2x}=e^{-x^2(1-\frac 2x)}\le e^{-\frac{x^2}2}\to 0$

Because when $x\gg 1$ then $1-\frac 2x\ge \frac 12$

Answer 4

Firstly, this is same as $e^{-(x-1)^2+1}$. So it suffices to show that $e^{-(x-1)^2}$ converges to $0$. You can substitute $y=x-1$ noting that $x \to \infty $ iff $y\to \infty$.

So it suffices to show that $e^{y^2}$ diverges to infinity. You can observe for any $m>1$, every $y>m$ satisfies $$e^{y^2}>y^2>m^2>m$$ Basically this is equivalent to saying $e^{y^2} \to \infty$.

Answer 5

Let $\epsilon > 0$ be given. We must find a $\beta$ such that $e^{-x^2+2x} < \epsilon$ whenever $x>\beta.$ Let $$\beta = \sqrt{\log(1/\epsilon) + 1}+1.$$ Then for $x>\beta,$ $$ x-1> \sqrt{\log(1/\epsilon)+1}\\(x-1)^2 > \log(1/\epsilon )+1\\ x^2-2x>\log(1/\epsilon)\\e^{-x^2+2x} <\epsilon$$

Answer 6

You need to deal with some inequalities. The given expression can be written as $\exp(2x-x^{2})$ and note that $2x-x^{2}<-x$ if $x>3$. And therefore if $x>3$ then the given expression is less than $e^{-x}=1/e^{x}$. Using the fundamental inequality $e^{t} \geq 1+t$ we can see that the given expression is less than $1/(1+x)$ and therefore less than $1/x$. Thus we have $$x>3\implies f(x) = e^{-x^{2}}e^{2x}<\frac{1}{x}$$ and the RHS can be made less than any positive $\epsilon$ if $x>1/\epsilon$. And thus we can see that if $\epsilon >0$ then we can choose $N=\max(3,1/\epsilon)$ such that $|f(x) |<\epsilon$ whenever $x>N$. It follows that $f(x) \to 0$ $x\to\infty$.

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It is not necessary/preferable to solve the inequality $|f(x) |<\epsilon$ and use logarithm in the process. The definition of limit is based on one way implication and solution of inequalities is based on two way implication and hence involves stricter procedures than necessary for establishing a limit.