I want to show that a function of the form $f(x)=\sqrt{x+a}+b$ with $a$ and $b$ some constants is, as $x\to\infty$, equal to $$ f(x)=\sqrt{x}+O(x^r) $$ for some $0<r<1/2$ (any such $r$ would be fine). Intuitively it seems to me that this should be true. Isn't it so? Or how can it be proved?

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    $\begingroup$ Write down $f(x) - \sqrt{x} = (\sqrt{x+a} - \sqrt{x}) + b$. What could you do with that? $\endgroup$ – Daniel Fischer Dec 6 '16 at 15:58
  • $\begingroup$ What about $b$. $\endgroup$ – hamam_Abdallah Dec 6 '16 at 16:21
  • $\begingroup$ Alright! $\sqrt{x+a}-\sqrt x$ goes to zero, so that the remainder term is bounded by a constant. I completely missed that... thanks! $\endgroup$ – Federico Dec 7 '16 at 23:11

The question has basically been answered in the comment by Daniel Fischer. I reproduce it here:

$$ \lim_{x\to\infty} (\sqrt{x+a}-\sqrt x)=\lim_{x\to\infty}\frac{(\sqrt{x+a}-\sqrt x)(\sqrt{x+a}+\sqrt x)}{\sqrt{x+a}+\sqrt x)} $$

$$ =\lim_{x\to\infty} \frac a {\sqrt{x+a}+\sqrt x}=0. $$

Therefore, $$ f(x)=\sqrt x+O(1). $$ (And we could even take $r=0$).

  • $\begingroup$ In the question we have $r>0$. $\endgroup$ – hamam_Abdallah Dec 8 '16 at 21:33

Just a Hint

Near $+\infty$,

$$f(x)=\sqrt{x}(1+\frac{a}{x})^\frac 12+b$$






  • $\begingroup$ Where did 1/4 come from? $\endgroup$ – Antonio Vargas Dec 6 '16 at 23:24
  • $\begingroup$ I have the same question. In the right hand side, as far as I can see, we only have $o(\sqrt x)$, or am I wrong? $\endgroup$ – Federico Dec 7 '16 at 19:01
  • $\begingroup$ @Federico What do you think now about the answer. $\endgroup$ – hamam_Abdallah Dec 8 '16 at 17:08
  • $\begingroup$ as $x \to \infty$ : $\sqrt{x+a} = \sqrt{x} + \mathcal{O}(x^{-1/2})$. I know it because $\sqrt{x+a} - \sqrt{x}= \int_x^{x+a}\frac12 t^{-1/2} dt = \mathcal{O}(x^{-1/2})$ $\endgroup$ – reuns Dec 8 '16 at 17:25

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