# Math Help - Proof

1. ## Proof

Working on practice problems.

Is anybody able to complete this?

Give an epsilon - delta proof of the limit fact

2. ## Re: Proof

Originally Posted by MathRaven
Working on practice problems.

Is anybody able to complete this?

Give an epsilon - delta proof of the limit fact

To prove \displaystyle \begin{align*} \lim_{x \to 1} (6x - 5) = 1 \end{align*} you need to show that \displaystyle \begin{align*} 0 < |x - 1| < \delta \implies |(6x - 5) - 1| < \epsilon \end{align*}.

\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 \begin{align*} \delta = \frac{\epsilon}{6} \end{align*}, we have

\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.