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

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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
 

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  • #2
SammyS
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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 .
 
  • #3
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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
 

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