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Show that the set $ \left\{\dfrac{1}{x^2-1}\mid x\in(0,1)\right\} $ is not bounded.

We should assume that it is bounded, then try to prove the opposite, but I don't know where to start.

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  • $\begingroup$ Welcome to math.SE! This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level. $\endgroup$
    – user37238
    Sep 18, 2015 at 12:39
  • $\begingroup$ $x^2-1 = (x-1)(x+1)$, so the function has a vertical asymptote at $1$ and is therefore unbounded on $(0,1)$. ${}\qquad{}$ $\endgroup$ Sep 18, 2015 at 14:39

2 Answers 2

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This is one way to do it:

  1. Let $M< 0$ be a candidate for a bound
  2. Solve $\frac{1}{x^2-1}=M-1$ (be careful with multiple solutions, make certain that one of them is in $(0,1)$)
  3. Conclude $M$ is not a bound after all
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    $\begingroup$ Actually, the function is bounded above (by $-1$) for $x\in(0,1)$. So your method won't work as stated. You need to prove there is no lower bound. $\endgroup$ Sep 18, 2015 at 13:54
  • $\begingroup$ @AkivaWeinbergercolumbus Fixed. $\endgroup$
    – Arthur
    Sep 18, 2015 at 15:49
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$\frac{1}{x^2 - 1} < 0$ for all $x \in (0, 1)$. Notice that the value of the expression becomes more and more negative as $x$ approaches 1. So, let $M < 0$ be arbitrary. Then all you need to do to show that $\frac{1}{x^2 - 1}$ is unbounded is find a special value of $x \in (0, 1)$ such that,

$$ \frac{1}{x^2 - 1} < M $$

Do you see how to proceed from there? (Hint: rearrange the terms to find $x$ in terms of $M$ and ensure that the value that you choose is in $(0, 1)$).

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