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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