# help with proving limits using Epsilon-Delta definition

by shirewolfe
Tags: definition, epsilondelta, limits, proving
 P: 3 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.
 P: 696 Use the triangle inequality perhaps, |x + 1| = |x - 1 + 2| <= |x - 1| + 2 < d + 2.
HW Helper
PF Gold
P: 12,016
 Quote by shirewolfe 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:
First of all, you have NOT found the limit, and certainly not shown that it is 2!
What you have done, is to make a GUESS at the limit value!!
(As it happens, you've made a correct guess, but that is irrelevant; it's still a guess).

What you are to do now, is:
1. Does my guess (2) satisfy the properties that a limit must have?
2. Let $$|x-1|<\delta$$
3. Then, $$|x^{2}+1-2|=|x^{2}-1|=|x-1||x+1|$$
4. Now, by assumption, $$|x-1|<\delta$$
Let us make a further assumption, that $$\delta<1$$
Then, $$|x+1|<2 (\delta<1)$$
And:
$$|x^{2}+1-2|<2\delta, (\delta<1)$$
5. Let $$\epsilon>0$$
If we are to have $$|x^{2}+1-2|<\epsilon$$ for all x satisfying $$|x-1|<\delta$$ it is sufficient if the following inequalities are simultaneously satisfied:
$$2\delta<\epsilon$$
$$\delta<1$$

Hence, setting $$\delta=min(\frac{\epsilon}{2},1)$$ suffices.

That is, we were able to show that our guess (2) satisfy the properties a limit must have.

P: 3

## help with proving limits using Epsilon-Delta definition

By 'finding' 2 through the substitution of 1 in f(x)=[(x^2)+1] I meant that it was a possible limit whose validity must be varified through the E-d definition. Thank you for correcting me however, to avoid my own future confusion.

Thanks for the helpful guidance and clarification.
 Sci Advisor HW Helper PF Gold P: 12,016 I was perhaps a bit too snappish on that point, don't bite me back, though..

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