I need to find

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



I tried writing the logarithms as products:


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!

  • $\begingroup$ You want $\lim_{x\to0}$ or $\lim_{n\to\infty}$? $\endgroup$ – Julián Aguirre Jul 20 '12 at 9:04
  • $\begingroup$ I need $lim_{x\rightarrow0}$. Could you point me in the right direction? $\endgroup$ – Grozav Alex Ioan Jul 20 '12 at 9:06
  • 1
    $\begingroup$ @GrozavAlexIoan $\log(1+x)=x+o(x)$, $x\to0$. $\endgroup$ – Andrew Jul 20 '12 at 9:08
  • $\begingroup$ I tried that. Unfortunately $e$ is not one of the possible answers. $\endgroup$ – Grozav Alex Ioan Jul 20 '12 at 9:16
  • $\begingroup$ @Andrew's hint works all right (and does not yield the limit e). $\endgroup$ – Did Jul 20 '12 at 9:21

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

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

  • $\begingroup$ There is an error. Check the link between the second equation and the third: in the third, no inverses should appear, so the limit is much simpler. $\endgroup$ – D. Thomine Jul 20 '12 at 11:18
  • $\begingroup$ If $x$ yields $0$ wouldn't the result be $ln(f(x))=\sum^n_{k=1}0+O(\sum^n_{k=1}0)$ meaning $f(x)=e^0$ => $\lim f(x)=1$ ? $\endgroup$ – Grozav Alex Ioan Jul 20 '12 at 11:35

(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.

  • $\begingroup$ Can you please explain what you did at the first part (how the $\frac{log}{x}$ appeared) and how you got the $(\frac{1}{1+x}+\frac{2}{1+2x}+\dots+\frac{n}{1+nx})$? Thank you very much! $\endgroup$ – Grozav Alex Ioan Jul 20 '12 at 9:34
  • $\begingroup$ I took the log on both sides - are you familiar with $\log a^b=b\log a$? Are you familiar with l'Hopital? Do you know that the derivative of $\log(1+kx)$ is $k/(1+kx)? $\endgroup$ – Gerry Myerson Jul 20 '12 at 9:40
  • $\begingroup$ I'm familiar with the logarithm properties and l'Hopital. I didn't know the derivative of $log(1+kx)$ though. This means the answer is $e^{\frac{n(n+1)}{2}}$m right? $\endgroup$ – Grozav Alex Ioan Jul 20 '12 at 9:47
  • $\begingroup$ Note that the term in my answer that comes out to $n(n+1)/2$ is in the denominator. $\endgroup$ – Gerry Myerson Jul 20 '12 at 10:48
  • $\begingroup$ @GerryMyerson Why is that term in the denominator? Isn't that term essentially your du, where u is equal to that sum with all the logs? $\endgroup$ – Mike Jul 20 '12 at 11:11

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