Result of the limit: $\lim_{x \to \infty} \sqrt{x + \sin(x)} - \sqrt{x}$? From my calculations, the limit of 
$\lim_{x \to \infty} \sqrt{x + \sin(x)} - \sqrt{x}$
Is undefined due to $sin(x)$ being a periodic function, but someone told me it should be zero. 
I was just wondering if someone could please confirm what the limit of this function is? Thanks
Corey :) 
 A: $$\sqrt{x+\sin x}-\sqrt{x}=\frac{(x+\sin x)-(x)}{\sqrt{x+\sin x}+\sqrt{x}}=\frac{\frac{\sin x}{\sqrt{x}}}{\sqrt{1+\frac{\sin x}{x}}+1}\to \frac{0}{\sqrt{1+0}+1}= 0$$
A: Despite the periodicity of $\sin x$ the following inequalites hold: $$0\leftarrow\sqrt{x-1}-\sqrt{x}\le\sqrt{x+\sin x}-\sqrt{x}\le \sqrt{x+1}-\sqrt{x}\to0,$$ therefore your limit is $0$ by the squeeze theorem.
A: computing  $$\int_x^{x+\sin x} \frac{1}{2\sqrt t} dt $$ in two ways we find that $$\sqrt{x+\sin x} - \sqrt x =\frac1{2\sqrt{x+k\sin x}} \text{ for some } 0 < k < 1.$$ now letting $x \to \infty$ gives that $$\lim_{x\to \infty}\sqrt{x+\sin x} - x = 0. $$
A: multiply numerator and denominator by $\sqrt{x+\sin(x)}+\sqrt{x}$
A: as $x \to \infty$ we have (by rationalizing the numerator)
$$
\sqrt{x+1} - \sqrt{x} \to 0
$$
substitute $x-1$ for $x$ to obtain
$$
\sqrt{x-1} - \sqrt{x} \to 0
$$
now use:
$$
\sqrt{x+1} - \sqrt{x} \ge \sqrt{x+\sin x} - \sqrt{x} \ge  \sqrt{x-1} - \sqrt{x} 
$$
A: It would be nice to be able to do this
$$\sqrt{x+\sin x}-\sqrt{x}= \sqrt{x+\sin x -x}=\sqrt{\sin x}=\dots$$
But this just isn't valid. It would be nice to be able to just remove the radical signs. Can you think of a way to do this ? (hint: think difference of squares)
A: or use the binomial theorem
$$
\sqrt{x+\sin x}-\sqrt{x} = \sqrt{x} \left(\sqrt{1 + \frac{\sin x}
{x}}-1\right) \\
= \sqrt{x}\left( \frac{\sin x}x + O(\frac1{x^2}) \right) \to 0
$$
A: I think the limit is $0$, since $sin(x)$ is bounded: At best, it is $1$ or $-1$, but if $x \rightarrow \infty$, a bounded number will look like a const; it won't influence the limit, so basically you have $\sqrt{x}- \sqrt{x}= 0$
