(xn) is Cauchy when abs(xn-xm) tends to 0 as m,n tend to infinity.
The answer is yes is and only if da limit of the sequence is a bounded function.The suficiency derives directly from the definition of the uniform convergence. The necesity follows from making n tend to infinity in |fn(x)|
When we say as a variable n tends to infinity, we mean as n gets very very large. For example. If we look at 1/n as n tend to infinity, then n gets very large and 1/n goes to zero.
As x tends towards 0 (from >0), log(x) tend to - infinity. As x tends to + infinity so does log (x), though at a much slower rate.
It would tend towards infinity
(xn) is Cauchy when abs(xn-xm) tends to 0 as m,n tend to infinity.
The answer is yes is and only if da limit of the sequence is a bounded function.The suficiency derives directly from the definition of the uniform convergence. The necesity follows from making n tend to infinity in |fn(x)|
Sine does not converge but oscillates. As a result sine does not tend to a limit as its argument tends to infinity. So sine(infinity) is not defined.
They tend to have prestige.
people tend to die in their upper 80's and 90's
When we say as a variable n tends to infinity, we mean as n gets very very large. For example. If we look at 1/n as n tend to infinity, then n gets very large and 1/n goes to zero.
As x tends towards 0 (from >0), log(x) tend to - infinity. As x tends to + infinity so does log (x), though at a much slower rate.
Upper
Upper Class Italians tend to be relatively Francophilic.
No, limit can tend to any finite number including 0. It is also possible that the limit does not tend to any finite value or approaches infinity. Example: The limit of x^2+5 tend to 6 as x approaches -1.
It would tend towards infinity
First off, it's theta, not thita. That said, it's going to tend towards zero, I believe.