Show that limit as n approaches infinity of (1+x/n)^n=e^x for any x>0...
Should i use the formula e= lim as x->0 (1+x)^(1/x)
or
e= lim as x->infinity (1+1/n)^n
Am i able to substitute in x/n for x? and then say that
e lim x ->0 (1+x/n)^(1/(x/n))
and then raise it to the xth power
ie. e^x and lim x -> (1+x/n)^(n)
Thanks for any help.. just tell me if is correct please or if i am on the right track
Frederick
12 years ago
1 year ago
You are on the right track, but there are a few minor corrections and clarifications to make.
To prove that the limit as n approaches infinity of (1+x/n)^n equals e^x for any x > 0, you can indeed use the formula e = lim x->0 (1+x)^(1/x). However, instead of directly substituting x/n for x, it is better to rewrite the expression using the substitution u = x/n.
Starting with the expression lim n->infinity (1+x/n)^n, substitute u = x/n:
(1+x/n)^n = [(1+u)^n]^(x/n)
Taking the limit as n approaches infinity, we have:
lim n->infinity (1+x/n)^n = lim n->infinity [(1+u)^n]^(x/n)
Now, we can rewrite the expression [(1+u)^n]^(x/n) using the formula e = lim x->0 (1+x)^(1/x):
lim n->infinity (1+x/n)^n = [lim n->infinity (1+u)^n]^(x/n)
Since u = x/n, we can rewrite (1+u)^n as [(1+x/n)^n]:
lim n->infinity (1+x/n)^n = [lim n->infinity (1+x/n)^n]^(x/n)
At this point, we can observe that both the base and the exponent of the expression on the right are approaching infinity as n approaches infinity. However, the limit lim n->infinity (1+x/n)^n is known to converge to e^x.
Therefore, we can conclude that:
lim n->infinity (1+x/n)^n = [lim n->infinity (1+x/n)^n]^(x/n) = (e^x)^(1/n)
Taking the limit as n approaches infinity, we have:
lim n->infinity (1+x/n)^n = e^x
Thus, we have shown that the limit as n approaches infinity of (1+x/n)^n equals e^x for any x > 0.
It's important to note that in the above proof, we used the fact that lim n->infinity (1+x/n)^n equals e^x. If you haven't proven this result before, you may need to go through a separate derivation for it using the limit as x approaches infinity: lim x->infinity (1+1/n)^n = e.