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Find the limit of a numerical sequence without the use of L'Hopital's rule. $$\lim _{n\to \infty }\left(\frac{n!\cdot \:n}{3\left(n+1\right)!-5n}\right)$$

I think you have to multiply the numerator and denominator for something.I do not know what to do here with the factorial.

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  • $\begingroup$ Try something! And then if it fails, tell us about it. $\endgroup$ – zhw. Nov 18 '15 at 20:18
  • $\begingroup$ I tried, but it does not work :( $\endgroup$ – do.this Nov 18 '15 at 20:19
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$L=lim_{n\to \infty }\left(\frac{n!\cdot \:n}{3\left(n+1\right)!-5n}\right)\Rightarrow L=lim_{n\to \infty }\left(\frac{\frac{n!}{(n+1)!}\cdot \:n}{3\frac{(n+1)}{(n+1)!}-\frac{5n}{(n+1)!}}\right)\Rightarrow L=lim_{n\to \infty }\left(\frac{\frac{n}{(n+1)}}{3-\frac{5}{(n-1)!(n+1)}}\right)$

$L=lim_{n\to \infty }\left(\frac{\frac{1}{(1+\frac{1}{n})}}{3-\frac{5}{(n-1)!(n+1)}}\right)$

$L=\frac{1}{3-0}=\frac{1}{3}$

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