# using epsilon delta notation to prove (3-4x)=7 and find the delta that corresponds to epsilon = .02

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### 1 Answer

Using `epsilon-delta` notation to prove that `3-4x=7` is true for `x=-1` is the same as finding the limit `lim_{x->-1}(3-4x}=7`.

This is the same as saying that for every `epsilon>0` then we need to find some `delta>0` so that if `|x-(-1)|=|x+1|<delta` then `|3-4x-7|=|-4-4x|=|(-4)(1+x)|=4|x+1|<epsilon` is true.

Let `epsilon>0`

Then let `0<delta<epsilon/4` ` `

Then for all `|x+1|<delta<epsilon/4`

we see that `4|x+1|<epsilon`

and so the limit has been proven.

In this case, for `epsilon=0.02` then let `delta<0.02/4=0.005`.

Notice the wording on this is very particular with the let and for all statements, and it is easy to get those messed up.