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Let $f : [2, \infty) \to [0,1)$ is a function with $$f(x) = \dfrac{\sqrt{4 + \left[\ln(1 + \sqrt{x^2 - 4})\right]^2} - 2}{x-2}$$ for all $x \in [2, \infty)$. Prove that, for all $x \in (2, \infty)$, the function satisfies $\limsup_{s \to x^+} f(s) < 1$.

asrida
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    This function is not well defined at $x=2$. – CyclotomicField Dec 18 '19 at 04:07
  • disregarding the point $x=2$ where the function is not defined the question is equivalent to show that the function is bounded by one, because it is continuous, and that it doesn't take this value in it domain – Masacroso Dec 18 '19 at 05:04

2 Answers2

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HINT: note that for $x>2$ $$ f(x)<1\iff \log(1+\sqrt{x^2-4})<\sqrt{x^2-4}\iff \log(1+y)<y $$ for $y>0$.

Masacroso
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  • this just prove that $1$ is the upper bound, but how to prove that supremum $f(x) < 1$ ? – asrida Dec 18 '19 at 05:41
  • @asrida you dont need to prove that the supremum of $f$ in it domain is less than $1$. If $f$ is continuous then $$\lim_{x\to c}f(x)=\limsup_{x\to c}f(x)=\liminf_{x\to c}f(x)=f(c)$$ – Masacroso Dec 18 '19 at 06:02
  • Oh i see, thank you very much – asrida Dec 18 '19 at 06:17
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If you consider the function $$f(x) = \dfrac{\sqrt{4 + \left[\ln(1 + \sqrt{x^2 - 4})\right]^2} - 2}{x-2}$$ the first problem is its behavior around $x=2$.

Composing Taylor series around $x=2^+$, we have $$f(x)=1-2 \sqrt{x-2}+\frac{11 }{3}(x-2)+O\left((x-2)^{3/2}\right)$$ $$\lim_{x\to 2^+} \, f(x)=1\qquad \text{and} \qquad\lim_{x\to 2^+} \, f'(x)=-\infty$$

On the other side, when $x$ is large,we have $$f(x) \sim \frac{\sqrt{4+\log ^2\left({x}\right)}-2}{x}+O\left(\frac{1}{x^2}\right)$$ $$\lim_{x\to \infty} \, f(x)=0^+\qquad \text{and} \qquad\lim_{x\to \infty} \, f'(x)=0^-$$

For sure, what remains to be proved is that $f'(x)$ does not show any $0$ but here I am stuck.

Claude Leibovici
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