Is this true for $n\geq 0$: $\lim\limits_{x\to \infty} e^{-x^2}\frac d {dx}\frac{T_n}{T_{n+1}}e^{x^2}=1 $?

Question

Let $T_n$ denote Chebyshev polynomials of the first kind, I w'd like to show the below limit if it is true which it's hold for few first of them then how do i show this if it is true

for $n\geq0$ :$$\lim_{x\to \infty} e^{-x^2}\frac{d}{dx}\frac{T_n}{T_{n+1}} e^{x^2} = 1 \text{ ?}$$

Thank you for any help

Answer 1

Hint: $$\exp(-x^2)\cdot\frac{d}{dx}(f(x)\cdot\exp(x^2))=2x\cdot f(x)+f'(x)$$

Extra Hint: $$\frac{T_n}{T_{n+1}}\sim\frac{1}{2x}$$

$$\frac{T_n}{T_{n+1}}=\frac{1}{2x}+\frac{2xT_n-T_{n+1}}{2xT_{n+1}}=\frac{1}{2x}-\frac{T_{n-1}}{2xT_{n+1}}$$