- #1
shirewolfe
- 3
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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.
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.
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.