Let $f: \mathbb{R} \to \mathbb{R}$ be a strictly increasing function. Prove or disprove that

$$\lim\limits_{x\to \infty} f(x)=\infty$$

It seems pretty obvious to me that the function has no upper limits and the function is ever increasing. Does that suffice as a prove? I think not. Please help me with the formal proof. Thank you.

  • 4
    $\begingroup$ It's not true.${}$ $\endgroup$ – Git Gud Feb 27 '15 at 14:28
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    $\begingroup$ There are numerous examples -- for instance, $x\mapsto \frac{x^2}{x^2+1}$, which is bounded by $1$ yet strictly increasing (on $\mathbb{R}_+$). $\endgroup$ – Clement C. Feb 27 '15 at 14:32
  • $\begingroup$ Thank you everyone. You helped me open the block that I was facing. Thank you very much. $\endgroup$ – Swadhin Feb 27 '15 at 14:34
  • $\begingroup$ It's just a confusion: strictly increasing doesn't imply unbounded; the converse is true though $\endgroup$ – Alex Feb 27 '15 at 14:56

It's not impossible to be strictly increasing, while remaining below a certain upper bound.

I'll illustrate this in the form of a discrete example:

$$0.9, 0.99, 0.999, 0.9999, \ldots$$

This sequence is strictly increasing but its terms will never exceed $1$.

Can you come up with a function $f\colon \mathbb{R} \to \mathbb{R}$ with a similar behavior?


Consider the function $x\to \tan^{-1}(x)$ (also referred to as $x\to \arctan(x)$)

It looks like

enter image description here


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