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?
$$\lim\limits_{x \to +\infty} \left(\frac{\left((x-1)^2-x\ln(x)\right)(x)!}{(x+2)!+7^x}\right)$$
 A: This should be the limit of a sequence, given that $x!$ is only defined for nonnegative integer $x$ (unless you use the Gamma function).
First step: divide numerator and denominator by $(x+2)!$, so you get
$$
\lim_{x \to +\infty}
  \frac{
    \dfrac{(x-1)^2}{(x+2)(x+1)}-\dfrac{x\ln(x)}{(x+2)(x+1)}
  }{
    1+\dfrac{7^x}{(x+2)!}
  }
$$
Now, prove that
$$
\lim_{x\to\infty}\dfrac{x\ln(x)}{(x+2)(x+1)}=0,
\qquad
\lim_{x\to\infty}\dfrac{7^x}{(x+2)!}=0
$$
A: 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.
A: Just look at this part of it: 
$\frac{(x-1)^2 x!}{(x+2)!}$ these are the largest terms top and bottom. This has limit $1$. It should now be clear how to do a rigorous proof, for example multiply top and bottom by $\frac{1}{(x+2)!}$ then all of the terms will have finite limits.
A: $\lim\limits_{x \to +\infty} \left(\frac{\left((x-1)^2-x\ln(x)\right)(x)!}{(x+2)!+7^x}\right)$
$\lim\limits_{x \to +\infty} \left(\frac{\left((x-1)^2-x\ln(x)\right)}{\frac{(x+2)!}{(x)!}+\frac{7^x}{(x)!}}\right)=\lim\limits_{x \to +\infty} \left(\frac{\left(x^2-2x+1-x\ln(x)\right)}{(x+1)(x+2)+\frac{7^x}{(x)!}}\right)$
$\lim\limits_{x \to +\infty} \left(\frac{\left(1-\frac{2}{x}+\frac{1}{x^2}-\frac{\ln x}{x}\right)}{(1+\frac{1}{x})(1+\frac{2}{x})+\frac{1}{x^2}\frac{7^x}{(x)!}}\right)$
$\lim\limits_{x \to +\infty}\frac{7^x}{(x)!}=0$ 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.
