## Proving limits using epison and delta

I will use a small 'g' to represent delta.for the limit as x->3 1/x+1 = 1/4 I am able to get to the point of the proof |x-3|/|x+1|< 4e. I must now find a number M such that 1/|x+1|< M so I can replace 1/|x+1| with M and proceed to take 'g' = 4e/M. This is where I am getting confused. The teaching from Dr Evans states there is no M such that 1/|x+1|< M for all x, but if |x-1|<1 (i.e. if 'g' =1), then I can show that 3<x+1<5so that 1/|x-1| < 1/3. Therefore we set M=1/3. I am not sure how to get 3< x+1< 5. Do I get that by using 0 < |x-3| < 'g' (1) --> 3< x-3 < 4 --> 3< x+1 < 5 so that 1/|x+1| < 1/3? Even if I came up with the inequality correct, I do not see how that results in 1/3. How do I arrive at 1/3 using this method? I do not understand the math that results in these inequalities resulting in 1/|x+1| < 1/3. I have been trying to figure this out for some time. Can someone please explain it to me?