To prove $\displaystyle \displaystyle \begin{align*} \lim_{x \to 1} (6x - 5) = 1 \end{align*}$ you need to show that $\displaystyle \displaystyle \begin{align*} 0 < |x - 1| < \delta \implies |(6x - 5) - 1| < \epsilon \end{align*}$.
$\displaystyle \displaystyle \begin{align*} |(6x - 5) - 1| &< \epsilon \\ |6x - 6| &< \epsilon \\ 6|x - 1| &< \epsilon \\ |x - 1| &< \frac{\epsilon}{6} \end{align*}$
So by letting $\displaystyle \displaystyle \begin{align*} \delta = \frac{\epsilon}{6} \end{align*}$, we have
$\displaystyle \displaystyle \begin{align*} |x - 1| &< \delta \\ |x - 1| &< \frac{\epsilon}{6} \\ 6|x - 1| &< \epsilon \\ |6x - 6| &< \epsilon \\ |(6x - 5) - 1| &< \epsilon \end{align*}$
as required. Q.E.D.