Simple Explanation needed

Question

I have been struggling with basics lately, so for this problem is-The value of $e^{-\infty}$ is 0 because $\frac{1}{e^{\infty}}=\frac{1}{\infty}=0$. Am I right ?

Answer 1

The graph actually never reaches zero, since there's an asymptote at y=0. The proper terminology would be that the limit as x approaches $-\infty$ is 0 (my apologies for writing it out, I'm currently on mobile)

Answer 2

Technically speaking, you can not evaluate a function on the "points" $+\infty$ and $-\infty$, nevertheless, for some functions, they are accumulation points of the domain, thus you can evaluate the limit. In your case you have:

$$ \lim_{x\longrightarrow +\infty} e^{-x}=\lim_{x\longrightarrow +\infty} \frac{1}{e^{x}}=0 $$ This is a standard limit, but You can check the result using the very definition of limit. Indeed:

$$ \lim_{x\longrightarrow +\infty}f(x)=c\stackrel{def}{\Longleftrightarrow} \forall M>0 \;\exists x_{0}>0 \,:\;|f(x)-c|<M \,\forall x>x_{0} $$ In your case, using the fact that $e^{-x}>0\,\forall x\in\mathbb{R}$, this becomes:

$$ |e^{-x}|<M\Longleftrightarrow e^{-x}<M\Longleftrightarrow -x<\log(M)\Longleftrightarrow x>\log\left(\frac{1}{M}\right) $$ and thus the limit is satisfied.