# Finding a sequence of nonnegative numbers with special property

Prove that for every $r\in \Bbb R$, there is a sequence of nonnegative numbers $\{h_k\}$ that converges to $0$, and $\dfrac { \sin (\frac {1}{h_k})}{\sqrt {h_k}}$ converges to $r$, where ${h_k}$ goes to $0$.

It is easy to make such a sequence for $r=0$, but I don't know how to prove the statement.

Is there any hint?

Thanks for your help.

## 3 Answers

HINT: For infinitely many values of $h_k$ (namely for $h_k=\frac{1}{2k\pi-\pi/2}$), the numerator $\sin\left(\frac{1}{h_k}\right)$ equals $-1$ and your expression tends to $-\infty$ as $k\to+\infty$. For other values of $h_k$ (namely for $h_k=\frac{1}{2k\pi+\pi/2}$), the numerator $\sin\left(\frac{1}{h_k}\right)$ equals $1$ and your expression tends to $+\infty$ as $k\to-\infty$. For the values $h_k=\frac{1}{2k\pi}$ the numerator and thus the expression always equal zero so they obviously tend to zero.

Now, for any $r$ find similar values of $h_k$ that make your expression equal to $r$, where $-\infty<r<+\infty$. Since the expressions always equals $r$ it will clearly tend to $r$. You will be able to find such $h_k$ between the values I gave in my first paragraph: I gave explicit values for $r=0$. For other values you will not find an algebraic expression for $h_k$, but you can show that such a value exists.

To find the $h_k$ for a given $r$, consider the function

$$f(x)=\frac{\sin\frac{1}{x}}{\sqrt x}$$

From what I wrote above, we see that the image of the interval

$$\left(\frac{1}{2k\pi+\pi/2},\frac{1}{2k\pi-\pi/2}\right)$$

approaches all real numbers. Therefore for $k$ large enough there exists an $h_k$ in that interval such that $f(h_k)=r$. Those $h_k$ satisfy your requirements.

• yes I knew what you say, but I need some more help. – user115608 Nov 20 '15 at 15:51

For integer $k>r^2/2\pi$, let $h_k=(2\pi k +\arcsin (r/\sqrt {2\pi k}))^{-1}$ where the $\arcsin$ function takes values in $[-\pi/2,\pi/2]$.Then $$\frac {\sin (1/h_k)}{\sqrt {h_k}}=\frac {r}{\sqrt { 2\pi k}}.\frac {1}{\sqrt {h_k}}=r\left( 1+\frac {\arcsin (r/\sqrt {2\pi k}) }{2\pi k}\right)^{(-1/2)}.$$

For $x>0,$ let $f(x) = [\sin (1/x))]/\sqrt x.$ Then $f$ is continuous on $(0,\infty).$ Claim: For any $b>0,$ $f((0,b)) = \mathbb R.$ Why? We know there is sequence decreasing to $0$ along which $f \to \infty,$ and one along which $f \to -\infty.$ The interval $(0,b)$ captures the tail end of both sequences. Continuity and the intermediate value theorem then give the claim.

Let $r\in \mathbb R.$ Using the claim, we see that for each $n,$ there exists $h_n \in (0/1/n)$ such that $f(h_n) = r.$ The sequence $h_n$ does the job.