Asymptotics of functions involving square roots

Question

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?

Answer 1

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 $r=0$).

Answer 2

Just a Hint

Near $+\infty$,

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

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

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

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

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

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