Question

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

Answer 1

Using `epsilon-delta` definition of limits:

`14-5x=4=>10-5x=0`

Let `f(x)=10-5x`

Thus:

`lim_(x->2)10-5x=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->a)f(x)=L`

This implies that:

`forallepsilon>0, existsdelta>0` such that

`0<|x-a|< delta=>|f(x)-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 =>|(10-5x)-0|< epsilon`

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

`|10-5x|< epsilon => |-5||x-2|< epsilon=> |x-2|< epsilon/5`

Now they look similar and we can see that `delta = epsilon/5`

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

So, lets choose `delta = epsilon/5` and plug it back into our delta function.

`0<|x-2|< epsilon/5`

`0<5|x-2|< epsilon`

`0<|5x-10|< epsilon => |f(x)-L|< epsilon`

That concludes our proof.