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 $$\lim_{y\to\infty} f^{-1}(y) = b$$
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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
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$\begingroup$ 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
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$\begingroup$ @John, your proposed counterexample is bounded above. $\endgroup$ – Lubin Jan 23 '17 at 18:21
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$\begingroup$ 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
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$\begingroup$ ya @lubin it was my bad i forgot to mention that $\endgroup$ – pRivat Jan 23 '17 at 19:45
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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.
