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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?

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

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    $\begingroup$ Is $x$ supposed to take only integer values in the above? $\endgroup$ – Clement C. Dec 29 '15 at 13:49
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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 $$ \lim_{x\to\infty}\dfrac{x\ln(x)}{(x+2)(x+1)}=0, \qquad \lim_{x\to\infty}\dfrac{7^x}{(x+2)!}=0 $$

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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.

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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.

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$\lim\limits_{x \to +\infty} \left(\frac{\left((x-1)^2-x\ln(x)\right)(x)!}{(x+2)!+7^x}\right)$
$\lim\limits_{x \to +\infty} \left(\frac{\left((x-1)^2-x\ln(x)\right)}{\frac{(x+2)!}{(x)!}+\frac{7^x}{(x)!}}\right)=\lim\limits_{x \to +\infty} \left(\frac{\left(x^2-2x+1-x\ln(x)\right)}{(x+1)(x+2)+\frac{7^x}{(x)!}}\right)$
$\lim\limits_{x \to +\infty} \left(\frac{\left(1-\frac{2}{x}+\frac{1}{x^2}-\frac{\ln x}{x}\right)}{(1+\frac{1}{x})(1+\frac{2}{x})+\frac{1}{x^2}\frac{7^x}{(x)!}}\right)$
$\lim\limits_{x \to +\infty}\frac{7^x}{(x)!}=0$ 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.

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  • $\begingroup$ i don't get your first step $\endgroup$ – user12 Dec 29 '15 at 14:20
  • $\begingroup$ I divided numerator and denominator by $(x)!$ $\endgroup$ – diya Dec 29 '15 at 14:22
  • $\begingroup$ in the last step instead, i can just compute that little limit because others parts tend to 0? $\endgroup$ – user12 Dec 29 '15 at 14:24

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