Using the precise definition of a limit (epsilon & delta)

[Cal124]

I'm trying to practise, precise definition of a limit (epsilon & delta)
Just to check I'm along the right lines here's a previous question to the one I'm stuck on

If epsilon > 0 then there is delta >0 ... All that introduction stuff, then
Lim x-> 2 (3x-1) =5
Hence
|x-2| < delta then |3x - 6| < epsilon
>> |x-2|< epsilon/2
>> delta = epsilon/2
>> epsilon = 3delta

On my other problem I have the same setup but f(x) = 4x^2 +2 & L=18
So
|4x^2 -16| < epsilon |x-2|<delta
I've tried /4 but
I'm just not sure how I can remove the squared from the left of epsilon
Any help would be great

 

[SammyS]

I'm trying to practise, precise definition of a limit (epsilon & delta)
Just to check I'm along the right lines here's a previous question to the one I'm stuck on

If epsilon > 0 then there is delta >0 ... All that introduction stuff, then
Lim x-> 2 (3x-1) =5
Hence
|x-2| < delta then |3x - 6| < epsilon
>> |x-2|< epsilon/2 $\quad \quad$ That should be epsilon/3
>> delta = epsilon/2
>> epsilon = 3delta

On my other problem I have the same setup but f(x) = 4x^2 +2 & L=18
So
|4x^2 -16| < epsilon |x-2|<delta
I've tried /4 but
I'm just not sure how I can remove the squared from the left of epsilon
Any help would be great

I should have mentioned the above typo.

Linear functions work out nicely. Quadratic functions require somewhat more effort.

I assume that for the limit involving 4² +2, you also have x → 2 .

 

[Cal124]

I should have mentioned the above typo.

Linear functions work out nicely. Quadratic functions require somewhat more effort.

I assume that for the limit involving 4² +2, you also have x → 2 .

Awh sorry missed that. Yeah it's
Lim. (4x^2 +2) = 18
X--> 2