# How do you find values of δ that correspond to ε=0.1, ε=0.05, and ε=.01 when finding the limit of (5x-7) as x approaches 2?

Mar 6, 2015

Make $\delta < \frac{\epsilon}{5}$.

We believe that ${\lim}_{x \rightarrow 2} \left(5 x - 7\right) = 3$.

To show this, we need to show that we can make $\left\mid \left(5 x - 7\right) - 3 \right\mid < \epsilon$ by making $\left\mid x - 2 \right\mid < \delta$.

We observe that:
$\left\mid \left(5 x - 7\right) - 3 \right\mid = \left\mid 5 x - 10 \right\mid = \left\mid 5 \left(x - 2\right) \right\mid = \left\mid 5 \right\mid \left\mid x - 2 \right\mid = 5 \left\mid x - 2 \right\mid$.

Makng $\left\mid \left(5 x - 7\right) - 3 \right\mid < \epsilon$ can be done by making $5 \left\mid x - 2 \right\mid < \epsilon$.

Which, in turn, can be done by making $\left\mid x - 2 \right\mid < \frac{\epsilon}{5}$. That is, by making $\delta = \frac{\epsilon}{5}$.

For $\epsilon = 0.1$, make $\delta = \frac{0.1}{5} = 0.02$.
Now if $\left\mid x - 2 \right\mid < \delta$, then $5 \left\mid x - 2 \right\mid$ must be < $5 \delta$. (If $a < b$, then $5 a < 5 b$.)
So we can be sure that $\left\mid \left(5 x - 7\right) - 3 \right\mid$ which is equal to $5 \left\mid x - 2 \right\mid$ must be < $5 \delta$.

That is $\left\mid \left(5 x - 7\right) - 3 \right\mid < 5 \left(0.02\right) = 0.1$

For $\epsilon = 0.05$, make $\delta = \frac{0.05}{5} = 0.01$.
Now if $\left\mid x - 2 \right\mid < \delta$, then $5 \left\mid x - 2 \right\mid$ must be < $5 \delta$. (If $a < b$, then $5 a < 5 b$.)
So we can be sure that $\left\mid \left(5 x - 7\right) - 3 \right\mid$ which is equal to $5 \left\mid x - 2 \right\mid$ must be < $5 \delta$.

That is $\left\mid \left(5 x - 7\right) - 3 \right\mid < 5 \left(0.01\right) = 0.05$

For $\epsilon = 0.01$, make $\delta = \frac{0.01}{5} = 0.002$.