lim (1 - 1/x)^(-x) as x->+infinty
= 0
can you please explain to me how you got that
I can.
if you let x = infinity. Then what is 1/x really equal to?
It's wrong though.
My solution? Please correct me then sir.
I think this is a special identity representing e, but let me check.
If my Cal I recall is correct, if the exponent was positive, it is the well-recognized lim that goes to e. The fact that exponent is is negative, I am leaning to agree with you it goes to 0. How did you reach your conclusion.
Well, You have something that is always smaller than one being raised to big negative numbers. It will certainly not tend toward 0 because the exponentiation will grow faster than the -1/x will shrink. However I think it does converge on e rather than go on to infinity.
It's 1-1, how do you get -1? And it's to the -x, meaning the equation is actually 1/(1-1/x) which is certainly zero.
No. It is \[(1-\frac{1}{x})^{-x}\] I agree that the fraction will tend toward 0, but the whole expression tends toward \(1^\infty\)
Actually I contend that inside the bracket goes to 1 because \[(1 \div \infty)\] tends to zero.
Oh my bad didn't see 1-1/x, even then the resultant inside is still zero. 0^infinity = 0.
it's not \(1-\infty\) it is 1- 0
God I need to stop failing and go to bed lol. I'm sorry polpak you're right.
But that is leaves us with \(\frac{1}{1^\infty}\) which is undetermined form and needs special handling with l'hopital.
Have to take the log to bring down the x out of the exponent, then take the derivative, etc. It's kind of a bear. It looks like it might be e though.
We need Lokisan for this one. We need someone who has read everything concerning the number or the limit e.
Heh. Well I checked with Alpha, and it's definitely e. Glad I don't have to do this for my homework though.
http://www.wolframalpha.com/input/?i=lim+as+x+goes+to+infinity+%281+-+1%2Fx%29^%28-x%29
Good job.
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