If I understand your question correctly, such a sequence is an = x cos(πx). It has neither an upper nor lower bound. It's divergent, but its limit is neither infinity nor negative infinity.
Chat with our AI personalities
(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