# 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 :)

• Did you try multiply by $\sqrt{ x+\sin(x) }+\sqrt{x}$? – Luis Felipe May 10 '15 at 13:07
• Maybe $\sqrt{x+ \sin(x)}-\sqrt{x}=\sqrt{x\left(1+ \frac{\sin(x)}{x}\right)}-\sqrt{x}$. Now I believe bounded function $\sin(x)$ divided by $f(x) = x$, where $\lim_{x\to \infty} f(x) = \infty$ is 0. Therefore you are left with $\sqrt{x (1 + 0)} - \sqrt{x}$. – quapka May 10 '15 at 13:10
• @quapka when you take limit, you must take it to all, so $\sqrt{\infty(1+0)}-\sqrt{\infty}$ – Luis Felipe May 10 '15 at 13:13
• @LuisFelipeVillavicencioLopez Sure, you are right. – quapka May 10 '15 at 13:16
• When a function $f$ contains a periodic function you cannot automatically assume it does not exist. But when we do say "it does not exist because it is periodic", it is because the function appears to approach at least $2$ distinct unique limits. In your question, you have a limit in the $\infty - \infty$(indeterminate) form so you cannot say that it does not exist because it is perodic. Examples: $\lim_{x\to \infty}\frac{\sin(x)}{x} \to 0$, $\lim_{x \to \infty}\sin(x) = DNE$ – user222031 May 10 '15 at 13:19

$$\sqrt{x+\sin x}-\sqrt{x}=\frac{(x+\sin x)-(x)}{\sqrt{x+\sin x}+\sqrt{x}}=\frac{\frac{\sin x}{\sqrt{x}}}{\sqrt{1+\frac{\sin x}{x}}+1}\to \frac{0}{\sqrt{1+0}+1}= 0$$

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.

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 $$\lim_{x\to \infty}\sqrt{x+\sin x} - x = 0.$$

multiply numerator and denominator by $\sqrt{x+\sin(x)}+\sqrt{x}$

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: $$\sqrt{x+1} - \sqrt{x} \ge \sqrt{x+\sin x} - \sqrt{x} \ge \sqrt{x-1} - \sqrt{x}$$

or use the binomial theorem $$\sqrt{x+\sin x}-\sqrt{x} = \sqrt{x} \left(\sqrt{1 + \frac{\sin x} {x}}-1\right) \\ = \sqrt{x}\left( \frac{\sin x}x + O(\frac1{x^2}) \right) \to 0$$

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)

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 $\sqrt{x}- \sqrt{x}= 0$

• This doesn't work in general. Compare to $(x+\sin(x))^2-x^2$ or even $(x+\sin(x))-x$ – Tim B. May 10 '15 at 13:14
• But in this case, it works. – Derenge May 10 '15 at 13:14
• What is a result worth if the proof is wrong? – Tim B. May 10 '15 at 13:15
• try improving your arguments in math – Luis Felipe May 10 '15 at 13:15