# How do you use the epsilon delta definition to prove that the limit of x/(6-x)=1 as x->3?

Jan 8, 2017

We can prove that

$\left\mid f \left(x\right) - 1 \right\mid < \epsilon$ for $\left\mid x - 3 \right\mid < {\delta}_{\epsilon}$

if ${\delta}_{\epsilon} = \frac{3 \epsilon}{2 - \epsilon}$

#### Explanation:

This would not be necessary as:

$f \left(x\right) = \frac{x}{6 - x}$

is continuous in $x = 3$, so:

${\lim}_{x \to 3} f \left(x\right) = f \left(3\right) = \frac{3}{6 - 3} = \frac{3}{3} = 1$

However, as an exercise, we can see that:

$\left\mid f \left(x\right) - 1 \right\mid = \left\mid \frac{x}{6 - x} - 1 \right\mid = \left\mid \frac{x - 6 + x}{6 - x} \right\mid = \frac{\left\mid 2 x - 6 \right\mid}{\left\mid 6 - x \right\mid}$

Now consider $x \in \left(3 - \delta , 3 + \delta\right)$ with $\delta > 0$ so that:

$\left\mid x - 3 \right\mid < \delta$.

We have:

$\left\mid 6 - x \right\mid = \left\mid 3 + 3 - x \right\mid \le \left\mid 3 \right\mid + \left\mid 3 - x \right\mid < 3 + \delta$

and

$\left\mid 2 x - 6 \right\mid = \left\mid 2 \left(x - 3\right) \right\mid = 2 \left\mid x - 3 \right\mid < 2 \delta$

Putting this two inequalities together:

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

This means that given $\epsilon < 0$ if we choose ${\delta}_{\epsilon}$ such that:

$\epsilon = \frac{2 {\delta}_{\epsilon}}{3 + {\delta}_{\epsilon}}$

that is:

${\delta}_{\epsilon} = \frac{3 \epsilon}{2 - \epsilon}$

we will have:

$\left\mid f \left(x\right) - 1 \right\mid < \epsilon$ for $\left\mid x - 3 \right\mid < {\delta}_{\epsilon}$