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## Main Question or Discussion Point

I am having trouble proving the limits of quadratic functions such as the following. (I used "E" to represent epsilon and "d" for delta)

lim [(x^2)+1]

x->1

I found the limit, L, to equal 2 and have proceeded through the following steps:

|f(x) - L| < E

| [(x^2)+1] - 1| < E

|[(x^2)-1]| < E

|x+1||x-1| < E

While I also know that 0 < |x-1| < d.

(I was thinking of finding the bounded open interval in which |x+1||x-1| = |[(x^2)-1]| and substituting the greastest figure of the interval, which would be greater than |x+1|, in place of |x+1| so that I would have z|x-1|< E where z is an identified numerical value. However, in problems like the one above, |x+1||x-1| = |[(x^2)-1]| within a seemingly unbounded interval.)

please help. your time and assistance is very much appreciated.

lim [(x^2)+1]

x->1

I found the limit, L, to equal 2 and have proceeded through the following steps:

|f(x) - L| < E

| [(x^2)+1] - 1| < E

|[(x^2)-1]| < E

|x+1||x-1| < E

While I also know that 0 < |x-1| < d.

__My question__is how do you find the numerical relationship between**|x+1||x-1|**and**|x-1|**so that I may find**d**in terms of**E**?(I was thinking of finding the bounded open interval in which |x+1||x-1| = |[(x^2)-1]| and substituting the greastest figure of the interval, which would be greater than |x+1|, in place of |x+1| so that I would have z|x-1|< E where z is an identified numerical value. However, in problems like the one above, |x+1||x-1| = |[(x^2)-1]| within a seemingly unbounded interval.)

please help. your time and assistance is very much appreciated.