Vertical asymptope (Calculus)?
find the vertical asymptote if it exist for f(x)= 3/e^x-1
Yes it I did mean f(x)= 3/e^x-1 if there is not a limit I want to know why,
- Wayne DeguManLv 74 weeks ago
Without brackets I don't know what you mean.
i.e. is it 3/(eˣ - 1)...?
or, 3/(eˣ) - 1...?
There is a lot of difference between them.
- PopeLv 74 weeks ago
Your update indicates that the function is to be defined just as you wrote it.
f(x) = 3/eˣ - 1
Equivalently, f(x) = 3e⁻ˣ - 1
With that definition, function f is continuous across all real numbers, in which case its graph cannot have a vertical asymptote.
Your question about a limit is not clear. Being continuous, the function has a limit at every real number. It also has a limit as x approaches infinity.
lim f(x) = -1
As x approaches negative infinity, f(x) approaches infinity, so there is no limit in that direction.
I see now that an anonymous poster called the function 3/(e^x)-1.
However, he/she interpreted it as 3/(eˣ - 1), which is not the same thing at all.
After having clarified your definition, it seems unlikely that you meant that, but if you did, then the graph would indeed have a vertical asymptote. It also would have two horizontal asymptotes.
- alexLv 71 month ago
Is that f(x)= 3/(e^x-1) ?
- Anonymous1 month ago
Depends if you mean 3/e^(x-1) or 3/e^x-1
3/e^(x-1) does not have a vertical asymptote as the denominator will neve equal 0
but 3/(e^x)-1 has a vertical asymptote when x=0 as e^0 = 1 and 1-1=0