I'm supposed to find the following limit, given that $u,v$ are roots of $t^2-2t+2=0$ and $\cot\theta =x+1$.
$$\displaystyle \lim_{\theta \to 0}\dfrac{\left((x+u)^{n}-(x+v)^{n}\right)}{u-v}$$
My Attempt:
$$\begin{aligned}\displaystyle \lim_{\theta \to 0}\dfrac{\left((x+u)^{n}-(x+v)^{n}\right)}{u-v}&=\displaystyle \lim_{\theta \to 0}\dfrac{\left((x+u)^{n}-(x+v)^{n}\right)}{(x+u)-(x+v)}\\ &=\lim_{\theta \to 0} (x+v)^{n-1}+(x+u)(x+v)^{n-2}+\ldots+(x+u)^{n-1}(x-v)\end{aligned}$$
I'm not sure how to proceed. On way could be to make an equation whose roots are $(x+u)$ and $(x+v)$ which would come in terms of $\theta$ and limit could be evaluated, but that is rather tedious. Any hints are appreciated. Thanks
$u,v$ are $1\pm i$
$x+u,x+v$ are $\cot\theta-1+1\pm i=\dfrac{e^{\pm i\theta}}{\sin\theta}$
$$\dfrac{(x+u)^n-(x+v)^n}{x+u-(x+v)}=\dfrac{2i\sin n\theta}{2i\sin^{n+1}\theta}$$
Now set $\theta\to0$
I really doubt the validity of the limit, though