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

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

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

I am aware that to do so, we must show
\$existsbeta 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

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>lndfrac{1}{epsilon}\$ and with \$epsilon<e\$ then \$x>sqrt{lndfrac{1}{epsilon}+1}+1\$