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How would you find this limit ?

$\lim_{n \to \infty}(1+\ln(1-\frac{1}{2n}))^n$

Thank you.

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It seems a homework question, isn't it? So, what have you tried so far? –  Pacciu Dec 18 '11 at 15:29
    
Well, I actually came across this question while studying a problem, but it is not really a question in the homework. –  user20010 Dec 18 '11 at 15:38
    
Note that we can see what the answer should be. By the power series expansion for $\ln(1+x)$, we have $1+\ln(1-1/(2n))=1-1/(2n)+O(1/n^2)$. Now use $(1+t/n)^n\approx e^t$. –  André Nicolas Dec 18 '11 at 15:45
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3 Answers

up vote 2 down vote accepted

We have $$\lim\limits_{n \to \infty}(1+\ln(1-\frac{1}{2n}))^n = \lim\limits_{n \to \infty}\left((1+\ln(1-\frac{1}{2n}))^{\frac1{\ln(1-\frac{1}{2n})}}\right)^{n\ln(1-\frac{1}{2n})}= e^{\lim\limits_{n\to\infty} n\ln(1-\frac{1}{2n}) }$$ provided the limit in the exponent exists. (Since $\lim\limits_{n \to \infty}\left((1+\ln(1-\frac{1}{2n}))^{\frac1{\ln(1-\frac{1}{2n})}}\right) = \lim\limits_{x\to 0} (1+x)^{\frac1x}=e$.)

So it only remains to compute the limit from the exponent. We get $$\lim\limits_{n\to\infty} n\ln(1-\frac{1}{2n}) = -\frac12 \lim\limits_{n\to\infty} \frac{\ln(1-\frac{1}{2n})}{-\frac1{2n}}=-\frac12$$ using $\lim\limits_{x\to 0}\frac{\ln(1+x)}x=1$.

So the original limit is $e^{-1/2}$.


A similar basic idea can be used in many computations of this type -- to rewrite this as limits containing expression $\left(1+\frac1{f(n)}\right)^{f(n)}$ where $f(n)\to\infty$ or $(1+g(n))^{\frac1{g(n)}}$ where $g(n)\to0$ and the compute the new limit in the exponent.


You can also use wolframalpha and click there on "show steps":

limit n to infty (1+ln(1-1/(2n)))^n

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Take the limit of the logarithm:

$$ \eqalign{ \lim_{n\rightarrow\infty } \ln [ 1+\ln (1-{1\over 2n})]^n &=\lim_{n\rightarrow\infty }{\ln [ 1+\ln (1-{1\over 2n})]\over 1/n}\cr &=\lim_{n\rightarrow\infty }{{1\over [ 1+\ln (1-{1\over 2n})]}\cdot{1\over 1-{1\over2n}}\cdot {1\over2n^2}\over -1/n^2}\cr &=\lim_{n\rightarrow\infty }{{1\over [ 1+\ln (1-{1\over 2n})]}\cdot{1\over 1-{1\over 2n}}\cdot {-1\over2 } }\cr &=-1/2. } $$ So the original limit is $e^{-1/2}$.

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If there is any limit of the form $$\lim_{x\rightarrow a}\left ( 1+f\left ( x \right ) \right )^{\frac{1}{g\left ( x \right )}}$$such that$$\lim_{x\rightarrow a}f\left ( x \right )=\lim_{x\rightarrow a}g\left ( x \right )=0$$ It is given by $$e^{\lim_{x\rightarrow 0}{\frac{f\left ( x \right )}{g\left ( x \right )}}}$$

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