# Limit Calculation - Sequences at infinity

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 $\dfrac{1-x^4}{1-x^3}=\dfrac{(1-x)(1+x+x^2+x^3)}{(1-x)(1+x+x^2)}$. – André Nicolas Nov 30 '12 at 19:05

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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the power of $(1-\frac1n)$ in denominator is $3$ or $2$. – lab bhattacharjee Nov 30 '12 at 18:12
I thought it was 2. Thank you – Amr Nov 30 '12 at 18:14
thanks you very much ! – yuta Dec 1 '12 at 8:10

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

So, the limit is $\frac 4 3$

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If you multiplied the numerator by $n^4$ the numerator should be $n^4-(n-1)^4$ – Amr Nov 30 '12 at 18:19
@Amr, thanks for your observation. Could you please verify? – lab bhattacharjee Nov 30 '12 at 18:22
I didnt understand. emm verify what – Amr Nov 30 '12 at 18:22

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

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