Epsilon-delta question with limits

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In summary, the conversation discusses how to prove that the limit of a function as x approaches 1 does not exist. The Bolzano-Cauchy criterion is mentioned as a possible method to use. The conversation also brings up the idea of using sequences to show that the limit does not exist. The expert suggests using two points, x' and x'', where x' is slightly to the left of 1 and x'' is slightly to the right of 1, and showing that the absolute value of the difference between the function values at these points is always greater than 0.5. Overall, the expert emphasizes the importance of understanding the Bolzano-Cauchy criterion in determining the existence of a limit.
  • #1
danago
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Let [tex]
f(x) = \left\{ {\begin{array}{*{20}c}
x & {x < 1} \\
{x + 1} & {x > 1} \\
\end{array}} \right.
[/tex], and let [tex]\varepsilon = 0.5[/tex]. Show that no possible [tex]\delta > 0[/tex] satisfies the following condition:

For all x:
[tex]0 < \left| {x - 1} \right| < \delta \quad \Rightarrow \quad \left| {f(x) - 2} \right| < 0.5[/tex].

That is, for each [tex]\delta > 0[/tex] show that there is a value of x such that:

[tex]
0 < \left| {x - 1} \right| < \delta \quad \Rightarrow \quad \left| {f(x) - 2} \right| \ge 0.5
[/tex]

This will show that [tex]\mathop {\lim }\limits_{x \to 2} f(x) \ne 2[/tex]


I started by finding the set of x values for which the function produces values between 1.5 and 2.5.

For x<1:
[tex]
\begin{array}{l}
\left| {f(x) - 2} \right| < 0.5 \Rightarrow \left| {x - 2} \right| < 0.5 \Rightarrow x \in (1.5,2.5) \\
(1.5,2.5) \cap ( - \infty ,1) = \emptyset \\
\end{array}
[/tex]

Therefore there are no values of x<1 for which f(x) is within 0.5 of 2.

For x>1:
[tex]
\begin{array}{l}
\left| {f(x) - 2} \right| < 0.5 \Rightarrow \left| {x - 1} \right| < 0.5 \Rightarrow x \in (0.5,1.5) \\
(0.5,1.5) \cap (1,\infty ) = (1,1.5) \\
\end{array}
[/tex]

Therefore, f(x) is within 0.5 units of 2 if and only if [tex]x \in (1,1.5)[/tex]. This means that the only possible value for delta is 0, and since [tex]\delta > 0[/tex], this value of 0 is not possible, thus making no possible value for delta.

Is that how i should go about proving such a limit does not exist? Are there flaws in what i have done?

Thanks in advance,
Dan.
 
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  • #2
are u required to use only delta epsylon to show that the lim of this function as x-->1 does not exist, because if not there is a nicer way using sequences to show this?
 
  • #3
I can show you a general way using sequences, and also incorporating delta and epsylon in it, to show that this limit does not exist. YOu might want to try to use the Bolcano-Cauchy criterion for the existence of the limit. Are u familiar with it?
 
  • #4
sutupidmath said:
I can show you a general way using sequences, and also incorporating delta and epsylon in it, to show that this limit does not exist. YOu might want to try to use the Bolcano-Cauchy criterion for the existence of the limit. Are u familiar with it?

No i can't say i am familiar with it :( I am only just about to begin my first year in an undergraduate course in engineering, and thought id try and get a head start in the calculus.
 
  • #5
well it basically says that a function f is said to have a limit A, as x-->a if and only if for every epsylon there exists some delta such that for any two points from its domain, that is x',x"
that satisfy the condition abs(x'-a)<delta and abs(x"-a)<delta , than the following is fullfilled:

abs( f(x')-f(x"))<epsylon.

so the idea to show that your function does not have a limit as x-->1 is to find any two points x', and x'' ( also sequences if you like to) and to show that

abs(f(x')-f(x")) will always be greater than epsylon =.5

such numbers would be x'=(1-1/n) where n is a natural number, and x"=(1+1/n)
look these two numbers are one at the right of 1 and the other at the left. Do you see how to go now?
 
  • #6
Obviously, if x< 1, there exist x arbitrarily close to 1, that is |x-1|< [itex]\delta[/itex], such that f(x)= x is close to 1 and |f(x)- 2| is close to 1 not < 0.5.
 
  • #7
yeah, x'=(1-1/n) is close to 1 and x'<1, so obviously

abs(x'-1)=abs(1/n)<delta, whatever you chose delta to be but on the contrary

abs(f(x')-2)=abs(-1) which is obviously greater than 0.5.

But you can use the bolchano-cauchy's criterion as well, as i explained above, and you will get to the same result, depends which one you like more!
 
  • #8
Because in these cases where u are given to prove that the limit of some function is not, say A, the method above will work, because it simply contradicts the deffinition of a limit given my Cauchy. But if you are asked to prove whether the limit of some function exists or not, in general withoug given to prove whether the limit of f is A, than the method above won't be of any help. SO, in those cases in order to determine whether the limit of that particular function exists or not as x-->a, you need to use the Bolcano-Cauchy's criterion for the existence of the limit, as i explained in my previous posts.
 

Related to Epsilon-delta question with limits

1. What is an epsilon-delta question with limits?

An epsilon-delta question with limits is a type of mathematical problem that deals with the concept of limits in calculus. It involves using the epsilon-delta definition of a limit to prove the behavior of a function as it approaches a certain value.

2. How do you solve an epsilon-delta question with limits?

To solve an epsilon-delta question with limits, you must first understand the definition of a limit and how it is related to the epsilon-delta definition. Then, you must use algebraic manipulation and logical reasoning to find a suitable delta value that satisfies the given epsilon value and proves the limit.

3. What are the key concepts involved in an epsilon-delta question with limits?

The key concepts involved in an epsilon-delta question with limits are the definition of a limit, the epsilon-delta definition of a limit, algebraic manipulation, and logical reasoning. It is also important to have a good understanding of calculus and the properties of functions.

4. Why are epsilon-delta questions with limits important?

Epsilon-delta questions with limits are important because they help us understand the behavior of functions and their limits. They also provide a rigorous mathematical framework for proving the existence of limits and determining their values. These concepts are crucial in many areas of mathematics, science, and engineering.

5. Are there any tips for solving epsilon-delta questions with limits?

Yes, there are a few tips for solving epsilon-delta questions with limits. First, make sure you have a strong understanding of the definition of a limit and the epsilon-delta definition. Then, try to visualize the problem and use geometric intuition to guide your solution. Finally, practice solving different types of problems to improve your skills and confidence.

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