rate of lim exp(x) ~? rate of lim exp(-x)

**BCox** · Jul2-09, 07:38 AM

Hello:
The actual functional analysis for my rate of convergence is a bit more complicated. But essentially the problem I have is knowing if the following is true:

lim exp(x) -> infinity as x->infinity
is at the same rate as
lim exp(-x) -> 0 as x-> infinity

??? Would really appreciate the actual answer. Thank you!!!!!!

**ImAnEngineer** · Jul2-09, 07:59 AM

Rate as in slope of a function?

if
f(x) = e^x
g(x) = e^-x

then
f'(x) = e^x
g'(x) = -e^-x

so
f'(x) / g'(x) = -1

So in absolute value, the rate is the same. Although one is positive and the other negative. Does this answer your question?

**BCox** · Jul5-09, 09:51 AM

Yes, it does.

What if my f(x) = e^x + cot y ... where my y is another variable for which cot y may diverge to positive infinity? I am still only interested in the rate for when x -> infinity.

What would g(x) look like for the rate at which the new f(x) approaches infinity at the same rate that g(x) approaches zero?

**Preno** · Jul5-09, 10:34 AM

Originally Posted by ImAnEngineer

f'(x) = e^x
g'(x) = -e^-x

so
f'(x) / g'(x) = -1

Um, no.

Originally Posted by BCox

What would g(x) look like for the rate at which the new f(x) approaches infinity at the same rate that g(x) approaches zero?

What is "the rate at which <something> approaches infinity"? I mean, I know what it is for one function to approach infinity faster than another, but what is it for one function to approach infinity faster than another function approaches zero?

**BCox** · Jul5-09, 05:12 PM

Originally Posted by Preno

Um, no.
What is "the rate at which <something> approaches infinity"? I mean, I know what it is for one function to approach infinity faster than another, but what is it for one function to approach infinity faster than another function approaches zero?

Yes, I looked at that more closely and realized that is wrong. The question behind the question is, what is the rate of the following at which the function below approaches zero:

lim (1/y) arccot [ -exp(x)/sin(y) - cot(y) ] ->0 as x->infinity

where y is (-pi,0)

**ImAnEngineer** · Jul5-09, 05:22 PM

what a nonsense i wrote, sorry