0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ yes I knew what you say, but I need some more help. $\endgroup$ – user115608 Nov 20 '15 at 15:51
2
$\begingroup$

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 $$\frac {\sin (1/h_k)}{\sqrt {h_k}}=\frac {r}{\sqrt { 2\pi k}}.\frac {1}{\sqrt {h_k}}=r\left( 1+\frac {\arcsin (r/\sqrt {2\pi k}) }{2\pi k}\right)^{(-1/2)}.$$

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.