# Epsilon-delta proof for a limit of a troublesome function

• thenthanable
In summary, the conversation discusses finding a suitable value for δ to make the equation |(1/(x2-1)) +1)| less than ε. The speaker suggests restricting |x| to be less than 1 and using the minimum value of 2(ε)0.5 for δ. However, there is a problem when working backwards, as the range of ε becomes infinite. To avoid this, the speaker suggests keeping x away from ±1 by restricting |x| to be less than 1/2.
thenthanable
limx→01/(x2-1)=-1

0<|x|<δ and |((1/(x2-1)) +1)<ε

Working on the second equation, I eventually got
|x|< (ε|x-1||x+1|)0.5
Suppose I restrict |x|<1
-1<x<1,
then
0<x+1<2 → |x+1|<2
-2<x-1<0 → 0<1-x<2 → |1-x|=|x-1|<2 (Is this step correct? In one of the textbooks, it says that we must always have x>0 which is important to conclude that for example, |x-1|<2 in the context of this question. Unfortunately, I don't see why we must conclude that x>0 because doesn't the modulus get rid of all negative values?)

So my choice for δ works out to be δ=min{1,2(ε)0.5}

However when I work backwards the problem arises.
Suppose I choose δ=1 → 2(ε)0.5 > 1 → ε>0.25

When I draw the graph, my δ-range of x=0 starts from the asymptote at x=-1 to the other asymptote x=1, and if this is the case, doesn't that mean the range ε about f(x)=-1 is infinite?

You have$$\left |\frac 1 {x^2-1}+1\right |=\left | \frac {1+x^2-1}{x^2-1}\right |= \left |\frac{x^2}{x^2-1}\right |$$which you want to make small. The numerator will be small as ##x\to 0## but if ##x## gets anywhere near ##\pm 1##, that denominator will make the fraction large. So you must keep ##x## away from ##\pm 1##. That is easy since you have ##x\to 0##. So start by restricting ##|x|<\frac 1 2##. Figure out the smallest ##|x^2-1|## can be and overestimate ##\frac{x^2}{|x^2-1|}## by putting that smallest value in the denominator. Then you will be ready to figure out ##\delta## to make it all less than ##\epsilon##.

## 1. What is an epsilon-delta proof?

An epsilon-delta proof is a rigorous mathematical technique used to prove the existence of a limit for a troublesome function. It involves defining boundaries around a specific point on a graph and showing that the function's values approach a specific limit as the distance from the point to these boundaries gets smaller and smaller.

## 2. Why is an epsilon-delta proof necessary?

Epsilon-delta proofs are necessary because they provide a rigorous and precise way to prove that a limit exists for a function that is otherwise troublesome or difficult to evaluate. This is particularly useful in advanced mathematical fields, such as calculus and analysis.

## 3. How does an epsilon-delta proof work?

An epsilon-delta proof works by setting up two variables, epsilon and delta, that represent the distance from a specific point to the function's boundaries. The proof then involves showing that as epsilon approaches 0, delta also approaches 0 and that the function's values approach a specific limit.

## 4. Can you give an example of an epsilon-delta proof?

An example of an epsilon-delta proof is the proof of the limit of the function f(x) = x^2 as x approaches 2. This proof involves setting epsilon = 0.01 and finding a corresponding delta value that would ensure that the function's values are within 0.01 of the limit, which in this case is 4.

## 5. Are there any limitations to an epsilon-delta proof?

While epsilon-delta proofs are a powerful and widely-used technique in mathematics, they do have limitations. They may not work for all types of functions, and in some cases, alternative methods may be required to prove the existence of a limit. Additionally, constructing an epsilon-delta proof can be a time-consuming and challenging process, even for experienced mathematicians.

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