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Can you help me calculate the following limit ?

$\lim_ {n \to \infty} \frac{ 1- (1-\frac{1}{n} )^4 }{1- (1-\frac{1}{n})^3 }$

Intuitively, I can see that the numerator decays to zero much faster than the denominator. But how can I show it formally? (I tried to divide by $ \frac{1}{n} )^4 $ and by $\frac{1}{n})^3$ but without any success.

Help?

Thanks!

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    It will be easier if you change notation. We are looking at $\dfrac{1-x^4}{1-x^3}=\dfrac{(1-x)(1+x+x^2+x^3)}{(1-x)(1+x+x^2)}$.2012-11-30

3 Answers 3

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Hint: Apply L'Hôpital's rule, since you have the form $\frac{0}{0}$.

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$\frac{ 1- (1-\frac{1}{n} )^4 }{1- (1-\frac{1}{n})^3 }=\frac{(n-1)^4-n^4}{\{(n-1)^3-n^3\}n}=\frac{-4n^3+6n^2-4n+1}{-3n^3+3n^2-n}=\frac{4+\frac1n()}{3+\frac1n()}$

So, the limit is $\frac 4 3$

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    I didnt understand. emm verify what2012-11-30
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Hint Divide the numerator and the denominator by $(1-1/n)$ to get:

$ \frac{ 1- (1-\frac{1}{n} )^4 }{1- (1-\frac{1}{n})^3 }= \frac {1+(1-1/n)+(1-1/n)^2+(1-1/n)^3} {1+(1-1/n)+(1-1/n)^2} $

Now it is easy to show that the limit is $4/3$

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    I thought it was 2. Thank you2012-11-30