So I have a question about the limit $$\lim_{x \to -\infty} e^{x^3}.$$ It is obvious that this limit goes to zero, however, how would one justify it formally? I know I can use the continuity of a function to say $$\lim_{x \to a}f(g(x))=f(\lim_{x \to a}g(x)),$$ when $f$ is continous and $\lim_{x \to a}g(x)$ exists. However, in this case $g(x)=x^3$ and so the limit does not exist as $\lim_{x \to a}g(x)=-\infty$. So I cannot use that, here I think squeeze theorem is necessary. For $x<-1$ we have $e^{x^3}<e^x$ and so since it is well know that $\lim_{x \to -\infty} e^{x}=0$ (proven from $\epsilon-\delta$). By sqeueeze theorem the limit is $0$. I feel like I am overcomplicating things but I do not see a more straighforward answer.
Thanks!
Let $y=-x\to \infty$ then
$$0\le e^{x^3}=\frac1{e^{y^3}}\le\frac1y\to0$$