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I need to find

$\lim_{x\rightarrow0}f(x)$ for the following function:

$f:(0,+\infty)$

$f(x)=[1+\ln(1+x)+\ln(1+2x)+\dots+\ln(1+nx)]^\frac{1}{x}$

I tried writing the logarithms as products:

$\lim_{x\rightarrow0}[1+\ln(1+x)(1+2x)\dots(1+nx)]^\frac{1}{x}$

and as a sum and nothing is getting me anywhere.

Also I know I have to use the formula: $\lim_{x\rightarrow0}(1+x)^\frac{1}{x}=e$

Can someone please help me?

Thank you very much!

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    ?? There are other numbers than 1 and$e$in this world, yaknow... By the way, did you try anything at all yourself, or are you waiting for a full solution to appear on this page?2012-07-20

2 Answers 2

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Here is the solution, take $\ln$ to both sides, gives,

$\ln \left( f \left( x \right) \right) ={\frac {\ln \left( 1+\sum _{ k=1}^{n}\ln \left( 1+kx \right) \right) }{x}} $

Using Taylor expansion of $\ln(1+t)= t+ O(t) $ at the point $t=0$ with $t = {\sum _{k=1}^{n}\ln \left( 1+kx \right) } $ yields

$ \ln \left( f \left( x \right) \right) ={\frac {\sum _{k=1}^{n}\ln \left( 1+kx \right) }{x}} + \frac{O\left(\left( {\sum _{k=1}^{n}\ln \left( 1+kx \right) } \right)^2\right)}{x} $

Taking the limit as x goes to $0$ to both sides of the above equation gives

$ \lim_{x->0}\ln(f(x))=\ln(\lim_{x->0} f(x) ) =\sum _{k=1}^{n}{k}^{} =\frac{n(n+1)}{2}\,.$

Exponentiating the last result, we get the answer

$ {\rm e}^{\frac{n(n+1)}{2}} $

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    If$x$yields $0$ wouldn't the result be $ln(f(x))=\sum^n_{k=1}0+O(\sum^n_{k=1}0)$ meaning $f(x)=e^0$ => $\lim f(x)=1$ ?2012-07-20
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(Edited in response to comment)

$\log f={\log\bigl(1+\log(1+x)+\log(1+2x)+\cdots+\log(1+nx)\bigr)\over x}$ By l'Hopital, the limit, if it exists, is the same as the limit of ${\left({1\over1+x}+{2\over1+2x}+\dots+{n\over1+nx}\right)\over\left(1+\log(1+x)+\log(1+2x)+\cdots+\log(1+nx)\right)}$ But now you can just set $x$ equal to zero.

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    @Mike, you're right, I'll edit.2012-07-20