1. ## Fact check

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

Provided $\displaystyle u(0)=0$

Here is how I decided upon that

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

Now let $\displaystyle \psi=u(x)$

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

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

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

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

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

Something seems wrong

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

2. Originally Posted by Mathstud28
Just to make sure, I am pretty sure that I am right but $\displaystyle \text{ As }x\to{0}\csc(u(x))\sim\frac{1}{u(x)}$

Provided $\displaystyle u(0)=0$

Here is how I decided upon that

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

Now let $\displaystyle \psi=u(x)$

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

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

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

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

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

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 $\displaystyle \frac{1}{x} + \frac{x}{6} + \frac{7 x^3}{360} + .....$ so

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

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 ......

3. Originally Posted by mr fantastic
The Laurent Series of cosec x is $\displaystyle \frac{1}{x} + \frac{x}{6} + \frac{7 x^3}{360} + .....$ so

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

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.

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

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

and if $\displaystyle \lim_{x\to{c}}|f(x)|$ converges then $\displaystyle \lim_{x\to{c}}f(x)$ 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 $\displaystyle \lim_{x\to\infty}|f(x)|=0$

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

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

5. Originally Posted by mr fantastic
Does it work if f(x) = 1/x, for example?
Yes..yes it does

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

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

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

Giving us the commonly known limit

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

and from this theroem

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

But for my second I thought of a counter example

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

So what are the restrictions?

6. Can I make this generalization?

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

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

7. Originally Posted by Mathstud28
Can I make this generalization?

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

that $\displaystyle -1\leq{f(\tau,\zeta,\theta)}\leq{1}$?
As long as $\displaystyle \zeta \in \mathbb{R}$ the cosine is bounded between -1 and 1.

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

The cosine function is not bounded in the complex plane.

8. Originally Posted by TheEmptySet
As long as $\displaystyle \zeta \in \mathbb{R}$ the cosine is bounded between -1 and 1.

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

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

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

and for odd parity

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

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

9. 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

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

and for odd parity

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

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...