<s>[INST] $\lim_{n \rightarrow \infty} \frac{1-(1-1/n)^4}{1-(1-1/n)^3}$ Find
$$\lim_{n \rightarrow \infty} \dfrac{1-\left(1-\dfrac{1}{n}\right)^4}{1-\left(1-\dfrac{1}{n}\right)^3}$$
I can't figure out why the limit is equal to $\dfrac{4}{3}$ because I take the limit of a quotient to be the quotient of their limits.
I'm taking that $\lim_{n \rightarrow \infty}1-\left(1-\frac{1}{n}\right)^4 = 0$ and likewise that $\lim_{n \rightarrow \infty}1-\left(1-\frac{1}{n}\right)^3 = 0$, which still gives me that the limit should be 0.
 [/INST] If you expand the binomials, you get $\dfrac{1-\left(1-\dfrac{1}{n}\right)^4}{1-\left(1-\dfrac{1}{n}\right)^3}=\dfrac{1-\left(1-\dfrac{4}{n}+\dfrac{6}{n^2}-\dfrac{4}{n^3}+\dfrac{1}{n^4}\right)}{1-\left(1-\dfrac{3}{n}+\dfrac{3}{n^2}-\dfrac{1}{n^3}\right)}=\dfrac{\dfrac{4}{n}-\dfrac{6}{n^2}+\dfrac{4}{n^3}-\dfrac{1}{n^4}}{\dfrac{3}{n}-\dfrac{3}{n^2}+\dfrac{1}{n^3}}$. Multiplying by $n$ on the top and bottom yields $\dfrac{4-\dfrac{6}{n}+\dfrac{4}{n^2}-\dfrac{1}{n^3}}{3-\dfrac{3}{n}+\dfrac{1}{n^2}}$. Then the limit is easier to evaluate. Any term with a division by $n$ will drop out as $n\to \infty$.
$\lim_{n \to \infty} \dfrac{1-\left(1-\dfrac{1}{n}\right)^4}{1-\left(1-\dfrac{1}{n}\right)^3}=\dfrac{4-0+0-0}{3-0+0}=\dfrac{4}{3}$
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