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?
$$\lim\limits_{x \to +\infty} \left(\frac{\left((x-1)^2-x\ln(x)\right)(x)!}{(x+2)!+7^x}\right)$$

I would do as following.
The limit equals $$\lim_{x \to \infty} \frac {(x - 1)^2 - x \log x} {x (x + 1) + 7^x/x!}.$$ Note that $(x - 1)^2 - x \log x \sim x^2$,  $x (x + 1) \sim x^2$ and $7^x/x! = o (1)$. Using these, we have $$\lim_{x \to \infty} \frac {(x - 1)^2 - x \log x} {x (x + 1) + 7^x/x!} = \lim_{x \to \infty} \frac {x^2} {x^2} = 1.$$ I know it's not rigorous but it's the way I process in my head when I glance at limits which look like complete mess like this one.
