How to compute $\lim\limits_{x \to +\infty} \left(\frac{\left((x-1)^2-x\ln(x)\right)(x)!}{(x+2)!+7^x}\right)$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it.
Can you explain the method and the steps used?
$$\lim\limits_{x \to +\infty} \left(\frac{\left((x-1)^2-x\ln(x)\right)(x)!}{(x+2)!+7^x}\right)$$

$\lim\limits_{x \to +\infty} \left(\frac{\left((x-1)^2-x\ln(x)\right)(x)!}{(x+2)!+7^x}\right)$
$\lim\limits_{x \to +\infty} \left(\frac{\left((x-1)^2-x\ln(x)\right)}{\frac{(x+2)!}{(x)!}+\frac{7^x}{(x)!}}\right)=\lim\limits_{x \to +\infty} \left(\frac{\left(x^2-2x+1-x\ln(x)\right)}{(x+1)(x+2)+\frac{7^x}{(x)!}}\right)$
$\lim\limits_{x \to +\infty} \left(\frac{\left(1-\frac{2}{x}+\frac{1}{x^2}-\frac{\ln x}{x}\right)}{(1+\frac{1}{x})(1+\frac{2}{x})+\frac{1}{x^2}\frac{7^x}{(x)!}}\right)$
$\lim\limits_{x \to +\infty}\frac{7^x}{(x)!}=0$ because factorial function grows faster than exponential functions.
Read Do factorials really grow faster than exponential functions?
$\lim\limits_{x \to +\infty}\frac{1}{x^2}=0$
$\lim\limits_{x \to +\infty}\frac{\ln x}{x}=0$
So our required limit is 1.
