Prove that the limit of |x|/x at x=0 DNE

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In summary: You are trying to prove the existence of a limit, but it looks like you are not following the correct steps.In order to prove the existence of a limit, the underlying condition is that for every ##\epsilon > 0##, there exists ##\delta > 0##. If you can show that this doesn't hold for a single value of ##\epsilon##, you are done. Also, you can observe that any ##\epsilon <1## works here, as the two sets of relations you can produce are always disjoint in that case.This is where the sequential definition of a limit is useful:$$\lim_{x \rightarrow a}f(x) = L$$ if
  • #1
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Homework Statement
.
Relevant Equations
.
Problem: Prove that $$\lim_{x\to 0}|\frac{|x|} x=\text{Undefined}$$

The solution written by my prof. uses a special case where the tolerance of error ##\epsilon=1/2##. However, I want to proof it with generality, meaning that the tolerance is shrinking.

Below is my attempt at a solution:
$$0<|x-0|<\delta;\, |f(x)-L|<\epsilon,\,\forall \epsilon>0$$
$$0<x^2<\delta^2;\, (f(0)-L)^2<\epsilon^2$$
$$\begin{cases}
f(x>0)=1\\
f(x<0)=-1
\end{cases}$$
This is where I am stuck. It looks the two part of the function are independent of ##\delta##, making it impossible to establish a relationship with ##\epsilon##. Can somebody please help?
 
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  • #2
Leo Liu said:
The solution written by my prof. uses a special case where the tolerance of error ##\epsilon=1/2##. However, I want to proof it with generality, meaning that the tolerance is shrinking.
Consider carefully what your professor's proof is doing. It says that there are points where the function values stay away from any proposed limit by at least 1/2. Doesn't that also work for any ##\epsilon \lt 1/2##?
 
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  • #3
FactChecker said:
Consider carefully what your professor's proof is doing. It says that there are points where the function values stay away from any proposed limit by at least 1/2. Doesn't that also work for any ##\epsilon \lt 1/2##?
She discusses the two parts separately and proves the limit DNE by arguing that there are two possible disjoint ranges in which L may lie.
 
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The only other way to do this that I can see is to consider the limits ##\lim_{x\to 0^{+}}f(x)## and ##\lim_{x\to 0^{-}}f(x)## and since they are not equal the limit does not exist, because if it existed they would be equal. I am afraid your professor's way is the only way using epsilon delta proof, the only thing that could change is the value of epsilon.
 
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  • #5
Leo Liu said:
Homework Statement:: .
Relevant Equations:: .

Problem: Prove that $$\lim_{x\to 0}|\frac{|x|} x=\text{Undefined}$$

The solution written by my prof. uses a special case where the tolerance of error ##\epsilon=1/2##. However, I want to proof it with generality, meaning that the tolerance is shrinking.

Below is my attempt at a solution:
$$0<|x-0|<\delta;\, |f(x)-L|<\epsilon,\,\forall \epsilon>0$$
$$0<x^2<\delta^2;\, (f(0)-L)^2<\epsilon^2$$
$$\begin{cases}
f(x>0)=1\\
f(x<0)=-1
\end{cases}$$
This is where I am stuck. It looks the two part of the function are independent of ##\delta##, making it impossible to establish a relationship with ##\epsilon##. Can somebody please help?
This is not mathematics. All I see are disconnected statements with no logical flow.
 
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  • #6
snip25.png

In order to prove the existence of a limit, the underlying condition is that for every ##\epsilon >0##, there exists ##\delta >0##. If you can show that this doesn't hold for a single value of ##\epsilon##, you are done. Also, you can observe that any ##\epsilon <1## works here, as the two sets of relations you can produce are always disjoint in that case.
 
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  • #7
This is where the sequential definition of a limit is useful:
$$\lim_{x \rightarrow a}f(x) = L$$ iff for every sequence ##\{x_n\}## in ##\mathbb{R}-\{a\}## we have $$\lim_{n \rightarrow \infty} x_n = a \ \Rightarrow \ \lim_{n \rightarrow \infty} f(x_n) = L$$
It's a good exercise to show this is equivalent to the ##\epsilon-\delta## definition.

In this case, you show that positive and negative sequences lead to different limits for the function values. This is similar to calculating the one-sided limits, as suggested above. But, in general, this sequential definition is more flexible (e.g. when you have things involving rationals and irrationals).
 
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  • #8
Say we note the function as f,
x ##\rightarrow## +0 f##\rightarrow## 1
x ##\rightarrow## -0 f##\rightarrow## -1
So
x ##\rightarrow## 0 f has no limit value.
f(x)=2H(x)-1
where H is Heaviside step function which has parameter of value H(0) as we like. Cf. https://en.wikipedia.org/wiki/Heaviside_step_function
 
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  • #9
PeroK said:
This is not mathematics. All I see are disconnected statements with no logical flow.
@Leo Liu If, for example, you are trying to show that a limit does not exist, then technically you need the negation of the limit. In this case:
$$\exists \epsilon > 0: \ \forall \delta > 0, \exists x: 0 < |x - a| < \delta \ \text{and} \ |f(x) - L| \ge \epsilon$$
This, however, says that a particular ##L## is not the limit. What we really want is to show that for all ##L \in \mathbb R##. Just for fun, let's do that anyway!:
$$\forall L \in \mathbb R: \exists \epsilon > 0: \ \forall \delta > 0, \exists x: 0 < |x - a| < \delta \ \text{and} \ |f(x) - L| \ge \epsilon$$
When you look at it like that it's no wonder you can get tangled up in the logic of showing formally that the limit does not exist. That's another reason to prefer the sequential approach or the one-sided limits. And, in general, it's a good idea to look for theorems that simplify what you need to show.

In any case, it's really important to have a logical flow to a proof and be crystal clear about what you are doing ar every step.
 
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  • #10
Thank you all!
 
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1. What does it mean for the limit of |x|/x at x=0 to not exist?

When we say that the limit of a function does not exist at a certain point, it means that the function does not approach a single, finite value as the input approaches that point. In this case, it means that as x approaches 0, the function |x|/x does not approach a single, finite value.

2. Why is it important to prove that the limit of |x|/x at x=0 does not exist?

Proving that the limit of |x|/x at x=0 does not exist is important because it helps us understand the behavior of the function at that point. It also allows us to identify any potential issues or discontinuities in the function, which can be important in further analysis or applications of the function.

3. How can we prove that the limit of |x|/x at x=0 does not exist?

To prove that the limit of |x|/x at x=0 does not exist, we can use the epsilon-delta definition of a limit. This involves showing that for any given value of epsilon (a small positive number), there does not exist a corresponding delta (a small positive number) such that all values of x within delta of 0 result in a function value within epsilon of a single, finite value.

4. What is the intuition behind why the limit of |x|/x at x=0 does not exist?

The limit of |x|/x at x=0 does not exist because the function approaches different values from the left and right sides of 0. From the left side, the function approaches -1, while from the right side, it approaches 1. This means that the function does not approach a single, finite value at x=0, and therefore the limit does not exist.

5. Are there any real-world applications of proving that the limit of |x|/x at x=0 does not exist?

Yes, there are real-world applications of proving that the limit of |x|/x at x=0 does not exist. For example, in physics and engineering, understanding the behavior of functions and their limits is important in predicting and analyzing physical phenomena. In this case, proving that the limit does not exist can help identify potential issues or discontinuities in a system or process.

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