I read about heavy right tail and I saw that a distribution is said to have a heavy right tail if its tail probabilities vanish slower than any exponential $$\forall >0: \lim_{x\to\infty}e^{tx}(>)=\infty.$$
Does the standard normal distribution have a heavy right tail?
The definition of a heavy right tailed distribution is that the moment generating function $M_X(t)$ is infinite for all $t > 0$ (see here). This is not the case for the standard normal distribution, where we have $$M_X(t)=\exp\left(\frac{t^2}{2}\right).$$ What you are writing about the limit $\lim_{x\to\infty}e^{tx}P(X>x)=\infty$ is only a necessary condition (an implication of the definition). But since it was part of your question: To calculate $\lim_{x\to\infty}e^{tx}P(X>x)$ you can use $P(X>x)=1-F_X(x)$ and then apply L'Hôpital's rule: $$\lim_{x\to\infty}e^{tx}P(X>x)=\lim_{x\to\infty}\frac{-\exp(-x^2/2)/\sqrt{2\pi}}{-t\cdot e^{-tx}}=0. $$ So by calculating this you also see that the standard normal distribution is not a heavy right tailed distribution.