A Big Limit

$$ Rewrite $$ $\begin{aligned} \lim_{x\to \infty} \left(\frac{x^2-5x+8}{x^2+1}\right)^{-3x+7}&=\lim_{x\to \infty} \left(1+\frac{7-5x}{x^2+1}\right)^{7-3x}\\ &=\lim_{x\to \infty} \left(1+\frac{(7-5x)(7-3x)}{(x^2+1)(7-3x)}\right)^{7-3x}\\ &=\lim_{x\to \infty} \left(1+\frac{\left(\frac{15x^2-56x+49}{x^2+1}\right)}{7-3x}\right)^{7-3x}.\\ \end{aligned}$ $$ By formal definition of exponential function , it is clear that $$ $\begin{aligned} \lim_{x\to \infty} \left(1+\frac{\left(\frac{15x^2-56x+49}{x^2+1}\right)}{7-3x}\right)^{7-3x}&=\exp\left(\lim_{x\to \infty}\frac{15x^2-56x+49}{x^2+1}\right).\\ \end{aligned}$ $$ Therefore $$ $\begin{aligned} N&=\exp\left(\lim_{x\to \infty}\frac{15x^2-56x+49}{x^2+1}\right)\\ &=\exp(15)\\ \lfloor N \rfloor&=3269017. \end{aligned}$ $$ Thus, the sum of the digits of $\,\lfloor N \rfloor=\boxed{28}$ . $$ $\text{\# }\mathbb{Q}.\mathbb{E}.\mathbb{D}.\text{\#}$