Im trying to solve this question:
let $f:[a,b)\rightarrow \mathbb{R}$ and $f$ is continuous and monotonically increasing and not bounded from above.
The first two questions were prove that $$\lim_{x\to b^-} f(x) = \infty$$
and: Prove that there is a function $g$ that for every $y\in[f(a),\infty)$ and for every $x\in[a,b)$ then $g(y)=x\iff f(x)=y$ (basically finding the "inverse" of the function if I understood it correctly).
Those two questions I managed to solve, but Im having trouble with the third one:
Prove that $$\lim_{y\to\infty} f^{-1}(y) = b$$

  • 1
    $\begingroup$ I think there is something wrong with your first question. Take $f(x)=x^2$ in the interval $[0,1)$. Then $f$ is clearly continuous and monotonically increasing. Yet $\lim_{x\rightarrow 1^-}x^2=1$. $\endgroup$ – John Jan 23 '17 at 17:55
  • $\begingroup$ What do you know about $f^{-1}$. What would it mean for that limit be less than b? greater than b? $\endgroup$ – lordoftheshadows Jan 23 '17 at 17:56
  • $\begingroup$ @John, your proposed counterexample is bounded above. $\endgroup$ – Lubin Jan 23 '17 at 18:21
  • $\begingroup$ Yes, but when I commented the question didn't mention that $f$ had to be unbounded. (look at the history) $\endgroup$ – John Jan 23 '17 at 19:06
  • $\begingroup$ ya @lubin it was my bad i forgot to mention that $\endgroup$ – pRivat Jan 23 '17 at 19:45

Call $f^{-1}=g$. Let $\varepsilon >0$ small enough, and call $M= f(b- \varepsilon )$. Then, $$y > M \ \Leftrightarrow \ g(y) > g(M) = b- \varepsilon \ \Leftrightarrow \ |b-g(y)| < \varepsilon $$

Hence $$\lim_{y \to \infty} g(y) = b$$ simply by direct check of the definition.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.