I am just wondering how to evaluate these limits. I am aware that the method I said is not mathematically acceptable as we do not have $\infty$ as powers etc. but I just want to see whether that make sense logically. I gave these limits and the procedure what I think and answers. Please comment whether I am right.
$$ \lim_{x\to\infty} e^{-ax} = \frac{1}{e^{a \infty}}=\frac{1}{\infty}=0 \\ \\ \lim_{x\to0} e^{-ax} = \big (e^{0} \big )^{-a} = \frac{1}{\big (e^{0} \big )^{a}}=1 $$
For every $a\in\mathbb{R}$, $$ \lim\limits_{x\to 0} e^{-ax}= e^{0} =1$$ If $a\lt 0$, then $$ \lim\limits_{x\to \infty} e^{-ax}= e^{\infty} =\infty $$ If $a=0$, then $$ \lim\limits_{x\to \infty} e^{-ax}= e^{0} =1 $$ If $a\gt 0$, then $$ \lim\limits_{x\to \infty} e^{-ax}= \frac{1}{e^{\infty}} =0 $$