Recent content by Duncan1

You are saying that \lim_{n \rightarrow \infty} (1 + 10^{-n}) = 1=lim_{n \rightarrow \infty} (1 - 10^{-n}) But I am saying that the limit is never actually reached, the expressions fall short by an infinitesimal amount. If you treat the above variations of 1 as interchangeable then I agree...

I disagree, if you do a binomial expansion of limn→∞ [(1+10-n)2·10n] = e2 The first term is 1^{+\infty} which you say is indeterminate, yet you seem happy to have a definite answer to this expression. Extrapolating n to increasing numbers does indicate convergence to a definite value. Further...

I disagree with 1^{+\infty} is indeterminate. If my initial equations are correct then 1^{+\infty} is bounded by e, 1/e. If the equations are changed to \lim_{n \rightarrow \infty} (1 + x(10^{-n}))^{10^n}=e^x Then for x=+1 and -1, 1^{+\infty} has a value between e^x and e^-x. limit as...

Thank you for trying to explain this to me. Clearly I am looking at the case where m=n for limits. What is the rule for separating limits? For example xsin(1/x) limit x->infinity or sin(y)/y limit y->0 would be undefined if you separate the limits. Or indeed my above examples would be...

However, \lim_{n \rightarrow \infty} (1 - 10^{-n})^{10^n}=1/e has a definite value other than 1. Try some values of n in a calculator. Is this expression in error? Does this expression not indicate that \lim_{n \rightarrow \infty} (1 - 10^{-n}) has a value other than 1 in the above...

Represent .999... by (1-10-n), limit as n>infinity. Then (1-10-n)10n, limit as n>infinity = 1/e (binomial expansion) (1+10-n)10n, limit as n>infinity = e (1)10n, limit as n>infinity=1 I think .999... is a transcendental number that can in almost every case be treated as equal to 1. Any...