Why does the limit of [(1/e^x)-1] / [(1/e^x)+1] equal -1 as x goes to infinity?

[YYaaSSeeRR]

I have a question and here it is :

[(1/e^x)-1] / [(1/e^x)+1]

why this equal -1 ?? when X →±∞I would appreciate it if you explain it for me on a paper after you capture it by your camera.

 

[Infrared]

Are you sure the answer is 1? The limit should be -1.

 

[YYaaSSeeRR]

sorry I forgot to put the - , yea it does equal -1 but why?

 

[Infrared]

What is [tex] \lim_{x\rightarrow \infty} {e^{-x}-1}?[/tex]
[tex]\lim_{x\rightarrow \infty} {e^{-x}+1} ?[/tex] Just divide these two answers.

 

[micromass]

What is [tex] \lim_{x\rightarrow \pm\infty} {e^{-x}-1}?[/tex]
[tex]\lim_{x\rightarrow \pm\infty} {e^{-x}+1} ?[/tex] Just divide these two answers.

That only works for the limit to ##+\infty##.

 

[YYaaSSeeRR]

That only works for the limit to ##+\infty##.

yes and that drives me crazy

 

[Infrared]

That only works for the limit to ##+\infty##.

Thanks, fixed.

For negative x, multiply the numerator and denominator by [itex] e^x [/itex]

 

[YYaaSSeeRR]

so what about when x→-∞ ?

 

[micromass]

so what about when x→-∞ ?

First things first. Do you understand the limit when ##x\rightarrow +\infty##?

 

[YYaaSSeeRR]

First things first. Do you understand the limit when ##x\rightarrow +\infty##?

yes I do.

 

[micromass]

yes I do.

Cool. For the other limit, you'll need to follow the hint in #7.

The answer won't be -1, by the way.

 

[YYaaSSeeRR]

For negative x, multiply the numerator and denominator by ex

I have not seen this after the modification :)
thanks a lot ,this problem forced me to throw the book away.

 

[micromass]

I have not seen this after the modification :)



thanks a lot ,this problem forced me to throw the book away.

So you found the right answer??

Also: you might want to post this in "calculus and beyond" next time

 

[YYaaSSeeRR]

So you found the right answer??

Also: you might want to post this in "calculus and beyond" next time

yes I got the right answer.

you must see teachers here in Syria ,they drive you crazy.
can't wait till I graduate high school and arrive to the US.

 

[YYaaSSeeRR]

actually when I followed hint #7 the answer wasn't -1 ,so Micromass what is the right answer?

 

[ChaseRLewis]

(e^-x - 1) / (e^-x +1)

Multiply num and denom by e^x

(1 - e^x) / (1 + e^x)

so you if you got positive 1 with negative infinite that makes sense as that reduces to
1/1 = 1

 

[YYaaSSeeRR]

(e^-x - 1) / (e^-x +1)

Multiply num and denom by e^x

(1 - e^x) / (1 + e^x)

so you if you got positive 1 with negative infinite that makes sense as that reduces to
1/1 = 1

(1 - e^x) / (1 + e^x) when x→-∞

yea it does equal 1 , but in my textbook when x→-∞ the equation (e^-x - 1) / (e^-x +1) equal -1 and that make no sense for me.

 

[micromass]

If your textbook says the answer is -1, then your textbook is wrong. The answer is 1.

 

[HallsofIvy]

Are we still talking about "[(1/e^x)-1] / [(1/e^x)+1]" as x goes to infinity? For very large x, 1/e^x is very close to 0 so the fraction is close to -1/1= -1.

(Oh, I see. The original post said "as [itex]x \to \pm\infty[/itex]" and the limit as x goes to negative infinity is 1.)

 

[micromass]

Are we still talking about "[(1/e^x)-1] / [(1/e^x)+1]" as x goes to infinity? For very large x, 1/e^x is very close to 0 so the fraction is close to -1/1= -1.

No, we're talking about the limit to ##-\infty##.