Does $\lim_{x\to\infty}f(x)e^{-x^2/2}=0$ imply $\lim_{x\to\infty}f'(x)e^{-x^2/2}=0$?

Question

I have a smooth function $f:\mathbb R\mapsto \mathbb R$ such that $$\lim_{x\to\infty}f(x)e^{-\frac{x^2}2}=0.$$

I wish to investigate under what condition(s) for $f$ if we also want $\lim_{x\to\infty}f'(x)e^{-\frac{x^2}2}=0.$

I have a conjecture that every analytic function that satisfies the first limit will do. To see this, we can use the fact that the derivative of analytic function is also analytic, so $\lim_{x\to\infty}f'(x)e^{-\frac{x^2}2}=0.$ Is my approach correct? Any suggestion?

Answer 1

Your conjecture is false, take for example $$ f(x) = \exp(x^2/2) \frac{\sin x^4}{1+x^2}. $$