- #1
MooCow
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Show that:
lim (1-1/n)^n=1/e
n->infinity
I don't really know where to begin...
lim (1-1/n)^n=1/e
n->infinity
I don't really know where to begin...
MooCow said:Show that:
lim (1-1/n)^n=1/e
n->infinity
I don't really know where to begin...
nicksauce said:Try taking the logarithm of both sides, then applying L'hopital's rule.
lurflurf said:That won't work because we are trying to establish the equality, thus we can not assume it to be true. Something similar works however.
let lim denote the limit n->infinity n a natural number
we wish to show
lim (1-1/n)^n=1/e
consider
log(lim (1-1/n)^n)=lim log((1-1/n)^n)
supose we know log is continuous at 1, then
log(lim (1-1/n)^n)=lim log((1-1/n)^n
perhaps we know log(x^y)=y*log(x), thus
log(lim (1-1/n)^n)=lim n*log((1-1/n)^n)
log(lim (1-1/n)^n)=lim -log(1-1/n)/(-1/n)
log(1)=0 so
log(lim (1-1/n)^n)=lim -[log(1-1/n)-log(1)]/(-1/n)
log(lim (1-1/n)^n)=-log'(1)
log(lim (1-1/n)^n)=-1
taking antilogs of both sides
lim (1-1/n)^n=exp(-1)=1/e
as desired
we do not need L'hopitals rule since we have the definition of the derivative
One serious problem you have with this is that it isn't true!MooCow said:Show that:
lim (1-1/n)^n=1/e
n->infinity
I don't really know where to begin...
HallsofIvy said:Except for that minor detail with the negative!
Ah, well, the operation was successful even though the patient died.
The limit (1-1/n)^n=1/e is an important concept in calculus and is often used in applications such as compound interest, population growth, and radioactive decay. It also has connections to the natural logarithm and the exponential function.
To prove this limit, we can use the definition of the limit and the properties of limits. First, we can rewrite the expression as lim (1-1/n)^n=lim (1-1/n)^n * 1=lim (1-1/n)^n * (n/n) = lim [(1-1/n)^n * (n/n)] = lim [(1-1/n)^n * (1/(1/n))] = lim [(1-1/n)^(n*n)] = lim (1-1/n)^n = lim (1-1/n)^n * lim (n/n) = lim (1-1/n)^n * lim (n/n) * lim (n/n) = lim (1-1/n)^n * lim (n/n) * lim (1/n) = lim [(1-1/n)^n * (1/n)] = lim (1-1/n)^n * (1/n) = lim (1-1/n)^n * lim (1/n) * lim (1/n) * lim (1/n) = lim (1-1/n)^n * lim (1/n) * lim (1/n) * lim (1/n) = lim (1-1/n)^n * lim (1/n) = 1 * 1 = 1/e. Therefore, lim (1-1/n)^n=1/e as n Approaches Infinity.
The limit (1-1/n)^n=1/e can be thought of as the continuous compounding of interest or growth over an infinite number of intervals. As n approaches infinity, the expression (1-1/n) gets closer and closer to 1, making the overall expression approach 1/e, which is the value of the exponential function at a rate of 1.
The natural logarithm, ln(x), is defined as the inverse function of the exponential function, e^x. Therefore, the value of ln(e) is equal to 1. By proving lim (1-1/n)^n=1/e, we are essentially showing that the limit of (1-1/n)^n approaches the value of e. This reinforces the concept that the natural logarithm is the inverse of the exponential function.
Yes, there are many real-world applications of this limit. For example, it can be used to model the growth of populations, the decay of radioactive materials, and the accumulation of interest in compound interest calculations. It is also used in finance, physics, and engineering to model various natural phenomena and make predictions.