$\lim_{n \rightarrow \infty} \frac{1-(1-1/n)^4}{1-(1-1/n)^3}$ 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.
 A: 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 $n\to \infty$.
$\lim_{n \to \infty} \dfrac{1-\left(1-\dfrac{1}{n}\right)^4}{1-\left(1-\dfrac{1}{n}\right)^3}=\dfrac{4-0+0-0}{3-0+0}=\dfrac{4}{3}$
A: $$\lim_{n \rightarrow \infty} \dfrac{1-\left(1-\dfrac{1}{n}\right)^4}{1-\left(1-\dfrac{1}{n}\right)^3}
\stackrel{\mathscr{L}}{=}\lim_{n \rightarrow \infty} \dfrac{4\left(1-\dfrac{1}{n}\right)^3 \dfrac{1}{n^2}}{3\left(1-\dfrac{1}{n}\right)^2\dfrac{1}{n^2}}
=\lim_{n \rightarrow \infty} \dfrac{4\left(1-\dfrac{1}{n}\right)^3}{3\left(1-\dfrac{1}{n}\right)^2} = \color{blue}{\dfrac{4}{3}}$$ 
A: Hint: $\dfrac{1-x^4}{1-x^3}= \dfrac{(1-x)(1 + x + x^2 + x^3))}{(1-x)(1+x+x^2)} = \dfrac{1 + x + x^2 + x^3}{1 + x+ x^2}$ (provided $x \neq 1$.)
Now put $x = 1 - \frac{1}{n}$ and take $n \to \infty.$
