# How do you prove that the limit of (3x − 5) = 1 as x approaches 2 using the epsilon delta proof?

Feb 16, 2017

GIven any $\epsilon > 0$ we choose ${\delta}_{\epsilon} < \frac{\epsilon}{3}$

#### Explanation:

We have to prove that given any $\epsilon > 0$ we can find a ${\delta}_{\epsilon} > 0$ such that:

$\left\mid f \left(x\right) - 1 \right\mid < \epsilon$ for $x \in \left(2 - {\delta}_{\epsilon} , 2 + {\delta}_{\epsilon}\right)$

We evaluate the difference:

$\left\mid f \left(x\right) - 1 \right\mid = \left\mid 3 x - 5 - 1 \right\mid = \left\mid 3 x - 6 \right\mid = 3 \left\mid x - 2 \right\mid$

and we can see that:

$\left\mid f \left(x\right) - 1 \right\mid < \epsilon \iff \left\mid x - 2 \right\mid < \frac{\epsilon}{3}$

So, given $\epsilon > 0$ we can choose ${\delta}_{\epsilon} < \frac{\epsilon}{3}$ and we have:

$x \in \left(2 - {\delta}_{\epsilon} , 2 + {\delta}_{\epsilon}\right) \implies \left\mid x - 2 \right\mid < \frac{\epsilon}{3} \iff \left\mid f \left(x\right) - 1 \right\mid < \epsilon$

which proves the point.