# How do you prove limit of 14-5x=4 as x->2 using the precise definition of a limit?

Dec 10, 2016

Using $\epsilon - \delta$ definition of limits:

$14 - 5 x = 4 \implies 10 - 5 x = 0$

Let $f \left(x\right) = 10 - 5 x$

Thus:

${\lim}_{x \to 2} 10 - 5 x = 0$

Algebraically, this makes sense; yet we want to prove this using the precise definition of a limit.

Since the general formula looks like:

${\lim}_{x \to a} f \left(x\right) = L$

This implies that:

$\forall \epsilon > 0 , \exists \delta > 0$ such that

$0 < | x - a | < \delta \implies | f \left(x\right) - L | < \epsilon$

This means that as we pick an interval on the x-axis that is close to $a$there will be an interval on the y-axis that is close to $L$.

So, if we plug in the values we know:

$0 < | x - 2 | < \delta \implies | \left(10 - 5 x\right) - 0 | < \epsilon$

We want to manipulate $| f \left(x\right) - L | < \epsilon$
so that it can represent $\delta$ as a function of $\epsilon$.

$| 10 - 5 x | < \epsilon \implies | - 5 | | x - 2 | < \epsilon \implies | x - 2 | < \frac{\epsilon}{5}$

Now they look similar and we can see that $\delta = \frac{\epsilon}{5}$

This gives us a ratio for when you're given a distance from $L$ (or an error tolerance).

So, lets choose $\delta = \frac{\epsilon}{5}$ and plug it back into our delta function.

$0 < | x - 2 | < \frac{\epsilon}{5}$

$0 < 5 | x - 2 | < \epsilon$

$0 < | 5 x - 10 | < \epsilon \implies | f \left(x\right) - L | < \epsilon$

That concludes our proof.