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Calculate the limit: $\lim \limits_{n \rightarrow \infty } \frac {4(n+3)!-n!}{n((n+2)!-(n-1)!)}$
My attempt: $\lim \limits_{n \rightarrow \infty } \frac {4(n+3)!-n!}{n((n+2)!-(n-1)!)} = \lim \limits_{n \rightarrow \infty } \frac {4(n+2)!(n+3)-n(n-1)!}{n((n+2)!-(n-1)!)} = \lim \limits_{n \rightarrow \infty } \frac {n(\frac{4}{n}(n+2)!(n+3)-(n-1)!}{n((n+2)!-(n-1)!)} = \lim \limits_{n \rightarrow \infty } \frac {\frac{4}{n}(n+2)!(n+3)-(n-1)!}{(n+2)!-(n-1)!} $
And here I am stuck.
I could do this: $\lim \limits_{n \rightarrow \infty } \frac {(4 +\frac{12}{n})(n+2)!-(n-1)!}{(n+2)!-(n-1)!}$But it doesn't get my anywhere. I was able to guess the right answer(4), but how to get to it in a proper calculation way?

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  • $\begingroup$ How about divide the numerator and the denominator by $(n-1)!$ first? $\endgroup$ – BAI Apr 8 '18 at 12:32
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    $\begingroup$ Or (quicker) divide both numerator and denominator by $(n+2)!$ $\endgroup$ – BAI Apr 8 '18 at 12:34
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We have \begin{align} \lim \limits_{n \rightarrow \infty } \frac {4(n+3)!-n!}{n((n+2)!-(n-1)!)} =& \lim \limits_{n \rightarrow \infty } \frac {(n+3)!\left[{4}-\frac{n!}{(n+4)!}\right]} {n(n+2)!\left[1-\frac{(n-1)!}{(n+2)!}\right]} \\ =& \lim \limits_{n \rightarrow \infty } \frac {(n+3)!\left[4-\frac{1}{(n+4)(n+3)(n+2)(n+1)}\right]}{n(n+2)!\left[1-\frac{1}{(n+2)(n+1)n}\right]} \\ =& \lim \limits_{n \rightarrow \infty } \frac{(n+3)}{n} \frac {\left[4-\frac{1}{(n+4)(n+3)(n+2)(n+1)}\right]}{\left[1-\frac{1}{(n+2)(n+1)n}\right]} \\ =& \lim \limits_{n \rightarrow \infty } \left(1+\frac{3}{n}\right) \frac {\left[4-\frac{1}{(n+4)(n+3)(n+2)(n+1)}\right]}{\left[1-\frac{1}{(n+2)(n+1)n}\right]} \\ =& \lim \limits_{n \rightarrow \infty } \left(1+\frac{3}{n}\right)\cdot \lim \limits_{n \rightarrow \infty } \frac {\left[ 4-\frac{1}{(n+4)(n+3)(n+2)(n+1)}\right]}{\left[1-\frac{1}{(n+2)(n+1)n}\right]} \\ =& \left(1+\lim \limits_{n \rightarrow \infty }\frac{3}{n}\right)\cdot \frac {\left[ 4-\lim \limits_{n \rightarrow \infty }\frac{1}{(n+4)(n+3)(n+2)(n+1)}\right]}{\left[1-\lim \limits_{n \rightarrow \infty }\frac{1}{(n+2)(n+1)n}\right]} \\ =& \left(1+0\right)\cdot \frac {\left[ 4-0\right]}{\left[1-0\right]}=4 \end{align}

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4
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Note that

$$ \frac {4(n+3)!-n!}{n((n+2)!-(n-1)!)}\sim \frac {4(n+3)!}{n((n+2)!)}= \frac {4(n+3)}{n}\to 4$$

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It is: $$\lim_\limits{n\to\infty} \frac{n!(4(n+1)(n+2)(n+3)-1}{n!(n(n+1)(n+2)-1)}=\\ \lim_\limits{n\to\infty} \frac{n^3\left[4\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\left(1+\frac{3}{n}\right)-\frac{1}{n^3}\right]}{n^3\left[\left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)-\frac{1}{n^3}\right]}=\frac{4(1+0)(1+0)(1+0)-0}{(1+0)(1+0)-0}=4.$$

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  • $\begingroup$ Have a question about factoring out from the denumerator n!. Are you sure that $n!*(1) = (n-1)!$ ? $\endgroup$ – trthhrtz Apr 8 '18 at 12:50
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    $\begingroup$ Sure, if you take $(n-1)!$ out of brackets and multiply by $n$, it becomes $n!$. $\endgroup$ – farruhota Apr 8 '18 at 12:53
  • $\begingroup$ Shouldn't in the denumerator be $n!(n(n+1)(n+2)-\frac{1}{n})$? $\endgroup$ – trthhrtz Apr 8 '18 at 12:56
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    $\begingroup$ Nope, it is: $n[(n-1)!\{n(n+1)(n+2)-1\}]=n(n-1)!\{n(n+1)(n+2)-1\}=n!\{n(n+1)(n+2)-1\}.$ $\endgroup$ – farruhota Apr 8 '18 at 13:03
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    $\begingroup$ I took the largest degree of $n$ (i.e. $n^3$) out of brackets (you can also divide by $n^3$). $\endgroup$ – farruhota Apr 8 '18 at 13:38
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You can collect $(n-1)!$ in the numerator and in the denominator, giving $$ \frac{(n-1)!}{(n-1)!}\frac{4(n+3)(n+2)(n+1)n-n}{n\bigl((n+2)(n+1)n-1\bigr)}= \frac{4n^4+f(n)}{n^4+g(n)}= \frac{4+\dfrac{f(n)}{n^4}}{1+\dfrac{g(n)}{n^4}} $$ where $f$ and $g$ are polynomials of degree less than $4$. Now use that $$ \lim_{n\to\infty}\frac{f(n)}{n^4}=\lim_{n\to\infty}\frac{g(n)}{n^4}=0 $$ There is no need to write down $f$ and $g$ explicitly.

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