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