- Thread starter Integrate
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\(\displaystyle \lim_{x->4}(3x+1) =13\) means that given any e>0 there is a d such that

|f(x) - 13|< e whenever 0<|x-4| < d

|3x+1-13|< e whenever 0<|x-4| < d

|3x- 12|< e whenever 0<|x-4| < d

3|x-4|< e whenever 0<|x-4| < d

|x-4|< e/3 whenever 0<|x-4| < d

Clearly that last statement above will be true if d = e/3. Do you see 100% why that last statement above will be true if d =e/3?? Will it be true if d>e/3? How about if d<e/3?

The goal is to get what is in green to be the same. If the left green is a multiple of |x-4| then you can divide by that multiple like I did above. Yes, you want to get the left side to be a multiple of the right side which is |x-4|

|f(x) - 13|< e whenever 0<|x-4| < d

|3x+1-13|< e whenever 0<|x-4| < d

|3x- 12|< e whenever 0<|x-4| < d

3|x-4|< e whenever 0<|x-4| < d

|x-4|< e/3 whenever 0<|x-4| < d

Clearly that last statement above will be true if d = e/3. Do you see 100% why that last statement above will be true if d =e/3?? Will it be true if d>e/3? How about if d<e/3?

The goal is to get what is in green to be the same. If the left green is a multiple of |x-4| then you can divide by that multiple like I did above. Yes, you want to get the left side to be a multiple of the right side which is |x-4|

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Perfect. It looks like that you understand it.oooooooh. So by the definition we need to to put the x->a into the template of 0<|x-a|<d.

we've more or less manipulated |f(x) - L| into exposing an |x-a|