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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 $x_n=1/n$.

$$f(x_n)=(1/n)^{1/n}=n^{-1/n}$$

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:

  1. Is there some $n$ beyond which $f$ is monotonically increasing?
  2. Can the subsequence be shown to be Cauchy? (algebraically tedious at best)
  3. 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

$$1/n<(1/n)^{1/n}<(1/n)^{1/m}<(1/m)^{1/m}$$

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.

$$|n^{-1/n}-m^{-1/m}|\le\ldots$$

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.

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  • $\begingroup$ To prove the limit, you must be able to show that the limit holds for EVERY subsequence going to 0, not just one. $\endgroup$
    – MT_
    Commented Feb 15, 2016 at 22:47
  • $\begingroup$ He wrote right at the beginning that the existence of the limit is assumed $\endgroup$
    – rafalpw
    Commented Feb 15, 2016 at 22:48
  • $\begingroup$ Limit existence was assumed in the problem, so if one subsequence has a limit there, every subsequence must. $\endgroup$
    – Kim Reece
    Commented Feb 15, 2016 at 22:48
  • $\begingroup$ Oh ok, makes sense! $\endgroup$
    – MT_
    Commented Feb 15, 2016 at 22:50

2 Answers 2

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IMHO the difficulty in answering this question is knowing what methods you are "allowed" to use. If the Sandwich (Pinching, Squeeze) Theorem is allowed,

  • $2^n>n^2$ for $n>4$ (easy induction proof)
  • so $0<\dfrac n{2^n}<\dfrac1n$;
  • so if you take $x_n=2^{-n}$ then $$2^{1/n}<x_n^{x_n}=2^{-n/2^n}<1$$ and the LHS tends to $1$ as $n\to\infty$.
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  • $\begingroup$ There's the bound I was missing! Yes, we are allowed to use Squeeze in this case, thank goodness. I can follow this approach through to a full solution. Thank you for the help. $\endgroup$
    – Kim Reece
    Commented Feb 16, 2016 at 3:08
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Hint: Let $x_n$ be a sequence converging towards $0$ $x_n^{x_n}=e^{x_nlog(x_n)}$. $lim_n x_nlog(x_n)=0$.

Limit of $x \log x$ as $x$ tends to $0$

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