Fact check

**Mathstud28** · June 10th 2008, 07:13 PM

Just to make sure, I am pretty sure that I am right but $\text{ As }x\to{0}\csc(u(x))\sim\frac{1}{u(x)}$

Provided $u(0)=0$

Here is how I decided upon that

$\lim_{x\to{0}}\frac{\csc(u(x))}{\frac{1}{u(x)}}=\l im_{u(x)\to{0}}\frac{u(x)}{\sin(u(x)}$

Now let $\psi=u(x)$

and as $x\to{0}\Rightarrow\psi\to{0}$

$\therefore\lim_{x\to{0}}\frac{u(x)}{\sin(u(x))}=\l im_{\psi\to{0}}\frac{\psi}{\sin(\psi)}=L$

so $\frac{1}{L}=\lim_{\psi\to{0}}\frac{\sin(\psi)}{\ps i}=\lim_{\psi\to{0}}\frac{\sin(\psi)-\sin(0)}{\psi-0}=\cos(0)=1$

So $\frac{1}{L}=1\Rightarrow{L=1}$

$\therefore\csc(u(x))\sim\frac{1}{u(x)}$

Something seems wrong

Could someone verify this? This is for something I don't want to make a factual error on. Thanks in advance

**mr fantastic** · June 10th 2008, 07:43 PM

Originally Posted by Mathstud28

Just to make sure, I am pretty sure that I am right but $\text{ As }x\to{0}\csc(u(x))\sim\frac{1}{u(x)}$

Provided $u(0)=0$

Here is how I decided upon that

$\lim_{x\to{0}}\frac{\csc(u(x))}{\frac{1}{u(x)}}=\l im_{u(x)\to{0}}\frac{u(x)}{\sin(u(x)}$

Now let $\psi=u(x)$

and as $x\to{0}\Rightarrow\psi\to{0}$

$\therefore\lim_{x\to{0}}\frac{u(x)}{\sin(u(x))}=\l im_{\psi\to{0}}\frac{\psi}{\sin(\psi)}=L$

so $\frac{1}{L}=\lim_{\psi\to{0}}\frac{\sin(\psi)}{\ps i}=\lim_{\psi\to{0}}\frac{\sin(\psi)-\sin(0)}{\psi-0}=\cos(0)=1$

So $\frac{1}{L}=1\Rightarrow{L=1}$

$\therefore\csc(u(x))\sim\frac{1}{u(x)}$

Something seems wrong

Could someone verify this? This is for something I don't want to make a factual error on. Thanks in advance

The Laurent Series of cosec x is $\frac{1}{x} + \frac{x}{6} + \frac{7 x^3}{360} + .....$ so

$\text{cosec} \, u = \frac{1}{u} + \frac{u}{6} + \frac{7 u^3}{360} + .....$

As to what happens when x --> 0, you obviously need to know something about the behaviour of u = u(x) .......

What you've got is valid I think when u ---> 0 as x --> 0 ......

**Mathstud28** · June 10th 2008, 07:47 PM

Originally Posted by mr fantastic

The Laurent Series of cosec x is $\frac{1}{x} + \frac{x}{6} + \frac{7 x^3}{360} + .....$ so

$\text{cosec} \, u = \frac{1}{u} + \frac{u}{6} + \frac{7 u^3}{360} + .....$

As to what happens when x --> 0, you obviously need to know something about the behaviour of u = u(x) .......

What you've got is valid I think when u ---> 0 as x --> 0 ......

Thanks a lot Mr. Fantastic, I am sure that if you think its right, it will be right enough for students of my age.

**Mathstud28** · June 10th 2008, 09:23 PM

Two more things, is it safe to say the following, are therea any caveats?

$\lim_{x\to\infty}f(x)^{\frac{1}{x}}=\lim_{x\to\inf ty}\frac{f(x+1)}{f(x)}$

and if $\lim_{x\to{c}}|f(x)|$ converges then $\lim_{x\to{c}}f(x)$ converges as well

I am pretty dang sure, but never hurts to check

Just to make sure I am not misuderstood

I know that

if $\lim_{x\to\infty}|f(x)|=0$

then $\lim_{x\to\infty}f(x)=0$

But I remember reading this assertion somewhere, I know it has pitfalls, but what are the restrictions

the more and more I think about it, the more I think its completely incorrect except at the few exceptions

**Mathstud28** · June 10th 2008, 09:44 PM

Originally Posted by mr fantastic

Does it work if f(x) = 1/x, for example?

Yes..yes it does

$\lim_{x\to\infty}\bigg(\frac{1}{x}\bigg)^{\frac{1} {x}}$

let $\xi=\frac{1}{x}$

Now as $x\to\infty\Rightarrow\xi\to{0}$

Giving us the commonly known limit

$\lim_{\xi\to{0}}\xi^{\xi}=1$

and from this theroem

$\lim_{x\to\infty}\bigg(\frac{1}{x}\bigg)^{\frac{1} {x}}=\lim_{x\to\infty}\frac{\frac{1}{x+1}}{\frac{1 }{x}}=\lim_{x\to\infty}\frac{x}{x+1}=1$

But for my second I thought of a counter example

$f(x)=(-1)^x$

So what are the restrictions?

**Mathstud28** · June 11th 2008, 08:56 PM

Can I make this generalization?

Let $f(\tau,\zeta,\theta)=\cos^{\tau}(\zeta\theta)$ where both $\tau$ and $\zeta$ are fixed constants, and $\tau\in\mathbb{Z}$

that $-1\leq{f(\tau,\zeta,\theta)}\leq{1}$ ?

**TheEmptySet** · June 11th 2008, 09:37 PM

Originally Posted by Mathstud28

Can I make this generalization?

Let $f(\tau,\zeta,\theta)=\cos^{\tau}(\zeta\theta)$ where both $\tau$ and $\zeta$ are fixed constants, and $\tau\in\mathbb{Z}$

that $-1\leq{f(\tau,\zeta,\theta)}\leq{1}$ ?

As long as $\zeta \in \mathbb{R}$ the cosine is bounded between -1 and 1.

$\cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}$

The cosine function is not bounded in the complex plane.

**Mathstud28** · June 11th 2008, 09:43 PM

Originally Posted by TheEmptySet

As long as $\zeta \in \mathbb{R}$ the cosine is bounded between -1 and 1.

$\cos(\theta)=\frac{e^{i\theta}+e^{-i\theta}}{2}$

The cosine function is not bounded in the complex plane.

Thanks EmptySet, But I should think you would know me a little better than that...I do know the bounds of cosine and its unboundedness in the complex plane, what I am saying is that if I said this in a formal setting would it be held as true, does the exponent change anything? I know that for a tau of even parity we have that

$0\leq\cos^{\tau}(\zeta\theta)\leq{1}$

and for odd parity

$-1\leq\cos^{\tau}(\zeta\theta)\leq{1}$

So can what I said be completely true, or is there a caveat I must include, I do not think so but I should be safe

**TheEmptySet** · June 11th 2008, 09:48 PM

Originally Posted by Mathstud28

Thanks EmptySet, But I should think you would know me a little better than that...I do know the bounds of cosine and its unboundedness in the complex plane, what I am saying is that if I said this in a formal setting would it be held as true, does the exponent change anything? I know that for a tau of even parity we have that

$0\leq\cos^{\tau}(\zeta\theta)\leq{1}$

and for odd parity

$-1\leq\cos^{\tau}(\zeta\theta)\leq{1}$

So can what I said be completely true, or is there a caveat I must include, I do not think so but I should be safe

I think that your analysis is correct. (But then again I may be crazy

)

Sorry I guess I missed the point of your question. It did seem odd for you to ask....
Rock On...

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