Anonymous
Anonymous asked in Science & MathematicsMathematics · 3 weeks ago

# Vertical asymptope (Calculus)?

find the vertical asymptote if it exist for f(x)= 3/e^x-1

Update:

Yes it I did mean f(x)= 3/e^x-1 if there is not a limit I want to know why,

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• 3 weeks ago

Without brackets I don't know what you mean.

i.e. is it 3/(eˣ - 1)...?

or, 3/(eˣ) - 1...?

or, 3/eˣ⁻¹...?

There is a lot of difference between them.

:)>​​​

• Pope
Lv 7
3 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

x→∞

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.

• alex
Lv 7
3 weeks ago

Is that f(x)= 3/(e^x-1) ?

• Anonymous
3 weeks 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