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I have to find the limit of the following:

$$\lim_{x\to\infty}{\frac{(x+1)(x^2+2)...(x^n+n)}{[(nx)^n+1]^{\frac{n+1}{2}}}}$$

I tried to find the logarithm of the upper side and lower side but it didn't work..what shall I do?

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Write down your expression here so we can read it, and it will be found when somebody comes by in 5 year's time would be a start. Look at the help for LaTeX imput (between $ inline, between $$ is display). –  vonbrand Feb 4 '13 at 11:40
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3 Answers

If you square the expression (call it $a(x))$ you get $$ a(x)^2=\frac{(x+1)^2(x^2+2)^2\cdot ...\cdot(x^n+n)^2}{(n^nx^n+1)^{n+1}}=\frac{x^{n(n+1)}+\ldots +n!^2}{n^{n(n+1)}x^{n(n+1)}+\ldots + 1} $$ $a(x)^2$ is a fraction of two polynomials in $x$, so we only need to look at the leading coefficients to conclude the limit of $a(x)^2$ is $n^{-n(n+1)}$. So the limit of $a(x)$ is $n^{-n(n+1)/2}$.

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Compare the greatest power of each side. The answer is $0$.

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Both numerator and denominator have $x^{n(n+1)/2}$, no? –  Gerry Myerson Feb 4 '13 at 11:43
    
Yes but not with the same coefficient ;) –  Damien L Feb 4 '13 at 11:44
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No,the answer is 1/[1+(1/n)]... –  Kyle92 Feb 4 '13 at 11:44
    
Damien, the coefficients are constants --- no one is taking a limit as $n\to\infty$. –  Gerry Myerson Feb 4 '13 at 12:24
    
Yes you are right. –  Damien L Feb 4 '13 at 12:41
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Hint: Think about $$\lim_{x\to +\infty}\frac{x^{1+2+3+...+n}}{(nx)^{\frac{n(n+1)}{2}}}$$

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I have thought about that,but it doesnt seem to tell me anything.. –  Kyle92 Feb 4 '13 at 11:46
    
@Kyle92: As it seems that, $n$ is a fixed constant. And the powers of variable $x$ in den. and num. are the same. –  B. S. Feb 4 '13 at 11:48
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