# Asymptotics of functions involving square roots

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?

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

## 2 Answers

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 $r=0$).

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

Just a Hint

Near $+\infty$,

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

$$=\sqrt{x}(1+\frac{a}{2x}(1+\epsilon(x))+b$$

$$=\sqrt{x}+\frac{a}{2\sqrt{x}}(1+\epsilon(x))+b$$

$$=\sqrt{x}+x^{\frac{1}{4}}(\frac{a}{2x^{\frac{3}{4}}}+\frac{b}{x^{\frac{1}{4}}})$$

$$=\sqrt{x}+x^{\frac{1}{4}}\epsilon(x)$$

$$=\sqrt{x}+O(x^{\frac{1}{4}}).$$

• Where did 1/4 come from? – Antonio Vargas Dec 6 '16 at 23:24
• 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? – Federico Dec 7 '16 at 19:01
• @Federico What do you think now about the answer. – hamam_Abdallah Dec 8 '16 at 17:08
• as $x \to \infty$ : $\sqrt{x+a} = \sqrt{x} + \mathcal{O}(x^{-1/2})$. I know it because $\sqrt{x+a} - \sqrt{x}= \int_x^{x+a}\frac12 t^{-1/2} dt = \mathcal{O}(x^{-1/2})$ – reuns Dec 8 '16 at 17:25