<s>[INST] Limit of Ratio of Chebyshev Polynomials I have been trying to compute the limit
$$\lim_{n\to\infty}{{U_n(x)^2}\over{U_{n-1}(x)^2+U_n(x)^2}}$$
where $U_n(x)$ is the $n$-th Chebyshev polynomial of the second kind and $x\ge 1$.
Using software I have been able to compute these limits exist when $x$ a half-integer, but I would like to have an explicit formula of the limit as a function of $x$ (at least for $x$ a half-integer). 
 [/INST] First thing we should note is:
$$\lim_{n\to\infty}\frac{U_n(x)^2}{U_{n-1}(x)^2+U_n(x)^2}=\lim_{n\to\infty}\frac{1}{\left(\dfrac{U_{n-1}}{U_n}\right)^2+1}$$
So, we look at $\displaystyle\lim_{n\to\infty}\frac{U_{n-1}}{U_n}$. Wikipedia gives the formula:
$$U_n=\frac{\left(x+\sqrt{x^2-1}\right)^{n+1}-\left(x-\sqrt{x^2-1}\right)^{n+1}}{2\sqrt{x^2-1}}$$.
Thus, the limit becomes:
$$\lim_{n\to\infty}\frac{\frac{\left(x+\sqrt{x^2-1}\right)^{n}-\left(x-\sqrt{x^2-1}\right)^{n}}{2\sqrt{x^2-1}}}{\frac{\left(x+\sqrt{x^2-1}\right)^{n+1}-\left(x-\sqrt{x^2-1}\right)^{n+1}}{2\sqrt{x^2-1}}}=\lim_{n\to\infty}\frac{\left(x+\sqrt{x^2-1}\right)^{n}-\left(x-\sqrt{x^2-1}\right)^{n}}{\left(x+\sqrt{x^2-1}\right)^{n+1}-\left(x-\sqrt{x^2-1}\right)^{n+1}}$$
Now let's call $a=x+\sqrt{x^2-1}$. Then $\frac{1}{a}=x-\sqrt{x^2-1}$, so the limits is now:
$$\lim_{n\to\infty}\frac{a^n-a^{-n}}{a^{n+1}-a^{-(n+1)}}$$
This depends on whether $a$ is $>$, $=$ or $<1$:
I'll do $|a|<1$ first:
We multiply the denominator and the numerator by $a^{n+1}$:
$$\lim_{n\to\infty}\frac{a^{2n+1}-a}{a^{2n}-1}$$
Here $a^{2n+1},a^{2n}\to0$ as $n\to\infty$, so the limit is equal to $a$.
In this case, the original limit is equal to $\frac{1}{a^2+1}=\frac{1}{2\left(x+\sqrt{x^2-1}\right)}=\frac{x-\sqrt{x^2-1}}{2}$.
The case where $|a|<1$ is very similar to this one.
The two remaining cases are $a=1$ and $a=-1$.
When $a=-1$, the denominator equals $2$ and the numerator is 0, so the limit does not exist and neither does the original one for values of $x$ for which $a=-1$, which is only $x=-1$.
The only case left is $a=1$ which gives $\frac{0}{0}$ which is undefined, so the limit is also undefined. This is for $x=1$.
</s>