How to prove that $\lim_{n \to \infty} \frac{3^n-1}{2 \cdot 3^n} = \frac{1}{2}$? I used this limit as an argument in a proof I wrote (Proof by induction that $\sum\limits_{k=1}^n \frac{1}{3^k}$ converges to $\frac{1}{2}$). 
I was told I should "prove" the limit but given no indication as to how to go about it. I didn't even know it was possible to formally prove a limit, but if it is I'd love to know how to do it.
 A: $$\lim_{n \to \infty} \frac{3^n-1}{2 \cdot 3^n} =\lim_{n \to \infty} \frac{\frac{3^n}{3^n} - \frac{1}{3^n}}{\frac{2\cdot 3^n}{3^n}} = \frac{1}{2}$$
A: This fraction can be written as a sum of two limits:
$$
\lim \frac{3^n-1}{2\cdot3^n}=\lim \frac{3^n}{2\cdot3^n}-\lim \frac{1}{2\cdot3^n}=\lim \frac{1}{2}-\lim \frac{1}{2\cdot3^n}=\frac{1}{2}-0=\frac{1}{2}
$$
A: \begin{equation*}
\frac{3^{n}-1}{2\cdot 3^{n}}=\frac{3^{n}}{2\cdot 3^{n}}-\frac{1}{2\cdot 3^{n}%
}=\frac{1}{2}-\frac{1/2}{3^{n}}
\end{equation*}
Since 
\begin{equation*}
\lim_{n\rightarrow \infty }\frac{1/2}{3^{n}}=0
\end{equation*}
then 
\begin{equation*}
\lim_{n\rightarrow \infty }\frac{3^{n}-1}{2\cdot 3^{n}}=\frac{1}{2}-0=\frac{1%
}{2}.
\end{equation*}
A: Split the limit up
$\lim_{x \rightarrow \infty} \dfrac{3^{n}}{2\cdot 3^{n}} - \dfrac{1}{2\cdot 3^n} = \dfrac{1}{2} - 0 $
It is pretty clear after splitting up the fraction.
A: $$\lim_{n\rightarrow\infty}\frac{3^n-1}{2\cdot 3^n}=\frac{1}{2}\lim_{n\rightarrow\infty}\frac{3^n}{3^n}-\frac{1}{3^n}=\frac{1}{2}(1-0)=\frac{1}{2}$$
A: Hi it's another way to calculate $S=\Sigma_{i=1}^{\infty}\frac{1}{3^i}$
Just multiply 3 at both side, then you will obtain
$3S=\Sigma_{i=1}^{\infty}3*\frac{1}{3^i} = 1+\Sigma_{i=1}^{\infty}\frac{1}{3^i}$
Do substraction, easy to get $S=\frac{1}{2}$
