$\displaystyle \lim_{n \to \infty} (\frac{3}{4})^n$
I know the answer is 0. But exactly how can I show that?
By prove it what tools do you mean? By delta-epsilon defintion?
We can prove that if $\displaystyle p>0,\alpha\in\mathbb{R}$ then $\displaystyle \lim_{n\to\infty}\frac{n^{\alpha}}{(1+p)^n}=0$
Proof: Consider an integerl $\displaystyle K>\alpha,K>0$. For $\displaystyle n>2K$
$\displaystyle (1+p)^n> {n\choose {K}} p^K>\frac{n^Kp^K}{2^KK!}$
This implies that
$\displaystyle 0\leqslant\frac{n^{\alpha}}{(1+p)^n}\leqslant\frac {2^KK!}{p^K}n^{\alpha-K}~~~n>2K$
Now you should know that if $\displaystyle \beta>0$ that $\displaystyle \lim_{n\to\infty}\frac{1}{n^{\beta}}=0$...so since $\displaystyle \alpha-K<0$ this proves that $\displaystyle \lim_{n\to\infty}\frac{n^{\alpha}}{(1+p)^n}=0$...take $\displaystyle \alpha=0,p=\frac{1}{3}$ to acheive your limit.
Is that what you wanted? Or were you looking for something simpler?
Note: I think if you weren't looking for a proof like the one above you could just state that $\displaystyle \left|\frac{3}{4}\right|<1$ so $\displaystyle \lim_{n\to\infty}\left(\frac{3}{4}\right)=0$