Growth of exponential functions with different base and/or exponent

Question

Assuming that the following is trivial:

1. $\lim_{x \to +\infty} \frac{2^x}{2.1^x}=0$

2. $\lim_{x \to +\infty} \frac{2^x}{2^{x^2}}=0$

What is the most simple, intuitive way to show that:

$\lim_{x \to +\infty} \frac{2^{x^2}2.1^x}{2.1^{x^2}}=0$

Answer 1

I think I got an explanation:

$2^{x^2}=({2^2})^{\frac{1}{2}x^2}=4^{\frac{1}{2}x^2}$

when $x \to +\infty$

$\frac{1}{2}x^2>x $

$\rightarrow\lim_{x \to +\infty} \frac{2.1^x}{4^{\frac{1}{2}x^2}}=0\rightarrow 2.1^x=o(2^{x^2})$

From this point it's really trivial (using $\lim_{x \to +\infty} \frac{2^{x^2}}{2.1^{x^2}}=0$)