Proving that $\lim_{x\rightarrow -\infty}e^x=0$ with epsilon delta definition

Question

I know that we need to find a way of relating $|x+\infty|<\delta$ and $|e^x-0|<\epsilon$. So, $|e^x-0|<\epsilon \iff e^x<e^{\log(\epsilon)}\iff x<\log(\epsilon)$. Then I imply that $\delta=\log(\epsilon)$ for $\epsilon>0$, but for $0<\epsilon<1$, $\delta$ is negative. How can I solve this proof?

Answer 1

The smaller $\epsilon$ is, the more negative $\log\epsilon $ is. So $x < \log \epsilon $ implies $x $ approaches $-\infty $ as $\epsilon $ approaches $0$, which is exactly what you want

Answer 2

The $\epsilon$-$\delta$ definition of limits doesn't apply here. You need a more general concept of limits, such as "Given any open set $S_1$ L, there is an open set $S_2$ 'containing' negative infinity such that $S_1$ contains the image of $S_2$". I've put "containing" in square quotes because infinity is not a real number, so a set of real numbers doesn't really contain it. However, we can extend the concept of "containing" to infinity to say that it's a limit. Thus, $S_1$ has to be in the form $(-\infty,x_0)$ for some $x_0$. Thus, we can give the condition we're trying to satisfy as $\forall \epsilon, \exists x_0: x \in (-\infty,x_0) \rightarrow |f(x)-L|<\epsilon$, or, equivalently, $\forall \epsilon, \exists x_0: x < x_0 \rightarrow |f(x)-L|<\epsilon$. For $L=0$, that is then $\forall \epsilon, \exists x_0: x < x_0 \rightarrow |f(x)|<\epsilon$. Since $f(x)$ is always positive, you can get rid of the absolute value. And since it's increasing, once you've proven that $f(x)< \epsilon$ for $x=x_0$, it follows that $f(x)< \epsilon$ for $x<x_0$. So now you just have to prove that for any $\epsilon$, there is some $x_0$ such that $f(x_0)=\epsilon$.

Answer 3

Let $y = -x$ and use the $(\epsilon, \delta)$ method to prove that $$ \lim_{y \rightarrow \infty} {1 \over e^y} = 0. $$

Answer 4

Since $-\infty$ isn't a number, the limit doesn't mean the same thing. Instead, it means "the limit of $e^x$ as $x$ decreases without bound." Put another way, the idea is that $e^x$ gets as close to $0$ as we like, so long as $x$ is sufficiently far to the left on the number line.

More precisely, we must show that for all $\epsilon>0,$ there is some real number $M$ such that $\bigl|e^x-0\bigr|<\epsilon$ whenever $x<M.$ Since you've already shown that $M=\log\epsilon$ does the trick, your proof is complete! Nicely done!

Answer 5

It is better to avoid inverse functions like $\log$ in $\epsilon, \delta$ proofs for limits dealing with $e^x$. Instead use the inequality $e^x\geq 1+x$.

Since $x\to-\infty$ we can assume $x<0$. Then we have $$e^x=\frac{1}{e^{-x}}\leq\frac{1}{1-x}<-\frac{1}{x}$$ and this is less than $\epsilon$ if $x<-1/\epsilon $. Thus we need to have $N=-1/\epsilon$ and then we have $$x<N\implies |e^x|<\epsilon $$ which means that $\lim_{x\to-\infty} e^x=0$.