Limit $\lim_{x \to \infty}{\sin{\sqrt{x+1}}-\sin{\sqrt{x}}}$

Question

I want to compute $$\lim_{x \to \infty}{\sin{\sqrt{x+1}}-\sin{\sqrt{x}}}.$$

Is it OK how I want to do?

$$\sin{\sqrt{x+1}}-\sin{\sqrt{x}}=2\sin{\frac{\sqrt{x+1}-\sqrt{x}}{2}}\cos{\frac{\sqrt{x+1}+\sqrt{x}}{2}}=2\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}}\cos{\frac{\sqrt{x+1}+\sqrt{x}}{2}}$$

I'm not surem but I want to say that $$|\cos \frac{\sqrt{x+1}-\sqrt{x}}{2}| \leq 1$$ $$\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}} \to 0 \mbox{ when } x \to \infty$$ So the limit is $0$ ?

I thinks is not because if I say that $ \displaystyle|\sin{\frac{1}{2(\sqrt{x+1}+\sqrt{x})}}| \leq 1$ I will obtain that another limit than $0$.

Thanks :)

Answer 1

Assuming you have the Mean Value Theorem at your dispoeal, that's the easiest way to show that the limit equals $0$: with $g(x) = \sin\sqrt x$, we have $$ \sin\sqrt{x+1}-\sin\sqrt x = g(x+1)-g(x) = ((x+1)-x)g'(\xi) = \frac{\cos\sqrt\xi}{2\sqrt\xi} $$ for some $x\le\xi\le x+1$. In particular, $$ |\sin\sqrt{x+1}-\sin\sqrt x| = \frac{|\cos\sqrt\xi|}{|2\sqrt\xi|} \le \frac1{2\sqrt x}, $$ and so $\lim_{x\to\infty} (\sin\sqrt{x+1}-\sin\sqrt x) = 0$ by the Squeeze Theorem.

Answer 2

Your approach is correct: The cosine part keeps oscillating, but it's bounded between -1 and 1, while the sine goes to zero due to its argument going to zero. Their product thus goes to zero too (even when multiplied by 2).

The final remark about bounding the value of sine is valid (in the sense of the inequality being true), but has no relevance -- if you know the sine and cosine are bounded between -1 and 1, you can conclude that their product falls between these bounds too (which is true, but doesn't provide much information about the actual limit or its existence).

Answer 3

A simpler argument is to write $f(x)=\sqrt{x+1}-\sqrt{x}$ and note that since $\lim\limits_{x\to \infty}f(x)=0$ and $\sin$ is uniformly continuous, we have $$\lim\limits_{x\to \infty}\sin(\sqrt{x}+f(x))-\sin(\sqrt{x})=0.$$

Why does this work? Let $g$ be a uniformly continuous function, $f$ a function such that $\lim\limits_{x\to\infty}f(x)=0$ and $h$ an arbitrary function. Then $$\lim\limits_{x\to\infty} g(h(x)+ f(x))-g(h(x))=0$$ as for any $\epsilon>0$ we can choose some $\delta>0$ such that $|a-b|<\delta\implies |g(a)-g(b)|<\epsilon$, so choosing $x_0$ such that $x>x_0\implies |f(x)|<\delta$ gives us that for $x>x_0$, $|g(h(x)+f(x))-g(h(x))|<\epsilon$, so the limit as $x\to \infty$ is $0$.

Answer 4

Here's another approach, using the OP's trick of getting the square roots into the denominator, but with a different trig identity, in this case $\sin(a+b)=\sin a\cos b+\cos a\sin b$.

$$\begin{align} |\sin(\sqrt{x+1})-\sin\sqrt x| &=|\sin(\sqrt x+\sqrt{x+1}-\sqrt x)-\sin\sqrt x|\\ &=|\sin\sqrt x ( \cos(\sqrt{x+1}-\sqrt x)-1)+\cos\sqrt x\sin(\sqrt{x-1}-\sqrt x)|\\ &\le|\cos(\sqrt{x+1}-\sqrt x)-1|+|\sin(\sqrt{x-1}-\sqrt x)|\\ &=\left|\cos\left(1\over\sqrt{x+1}+\sqrt x \right)-1\right|+\left|\sin\left(1\over\sqrt{x+1}+\sqrt x \right) \right|\\ &\to|\cos0-1|+|\sin0|=|1-1|+|0|=0 \end{align}$$