This is from Gaughan Ch.2, problem 10. The function $f(x)$ is defined from $(0, 2)$ into $R$. Limit existence is assumed. The hint is to choose a subsequence $x_n$ convergent to $0$ for which $f(x_n)$ is 'convenient'.
So far I have tried $x_n=1/n$ and $x_n=(1/2)^n$, with no luck in either case. Since the latter just ends up looping around to the former in the algebra, I'll explain here what I attempted to do with $x_n=1/n$.
We always have $0<1/n<1$, so any power or root of $1/n$ is necessarily likewise bounded. Therefore $f$ is bounded above by $1$ and below by $0$. Possible avenues of approach:
- Is there some $n$ beyond which $f$ is monotonically increasing?
- Can the subsequence be shown to be Cauchy? (algebraically tedious at best)
- Back to basics proof by definition of limit.
By graphical inspection, 1. does hold for $n\ge3$, although I have failed in all attempts to prove it. In fact, I get algebraically bogged down in about the same place trying either 1. or 2.
Take $n,m\in Z$ with $0<n<m$. Then $1/m<1/n$, so
At least half of which is useless; $1/n$ is going the wrong direction. The middle part isn't much use either, because it's in the middle.
All I can think is to go back to 1. and try induction. But I can't seem to factor the thing at all, so that's unclear. The only convergence herein demonstrated is between my head and the desk. Any suggestions would be appreciated.