# Thread: limit as t goes to negative infinity...

1. ## limit as t goes to negative infinity...

what is the limit of:

lim (t --> negative infinity) 1/e^t
?

because in this form the denominator shoots to negative infinity while the top remains 1, so it seems 0

but you can rewrite it as lim (t --> negative infinity) e^(-t)
where (-t) --> infinity as t --> negative infinity

and in this form we have an exponent shooting to infinity, so it seems like the answer is infinity

I ran into this problem late last night, so maybe I was just tired and didn't see something obvious, but I'm a bit confused about this, can anyone help?

2. Originally Posted by minivan15
because in this form the denominator shoots to negative infinity
e^t>0 and lim t-> -infinity e^t=0.

3. Originally Posted by treetheta
$\displaystyle \lim_{t\to-\infty} \frac{1}{e^t} = 0$
$\displaystyle \lim_{t\to-\infty} \frac{1}{e^t} =\lim_{t\to-\infty} {e^{-t}}= ?$
If $\displaystyle t=-100$ then what is $\displaystyle {e^{-t}}= ?$