Proving that 1/(x+1) is continuous at x=0 using epsilon-delta def.

  • Thread starter blackcat
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In summary, the person is trying to find a delta value so that |f(x)-1| < p and hence deduce that f(x) approaches 1 as x approaches 0. However, as x nears the point x=0, it can be positive or negative and he doesn't know how to get a delta value for this case. He asks for help and is given a way to approximate the delta value using limits.
  • #1
blackcat
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As the thread title. The question is actually:
"Given p > 0, find d so that |x-0| < d implies |f(x)-1| < p and hence deduce that f(x) approaches 1 as x approaches 0."

My problem is that, when x is near the point x=0, x can be positive and negative, I don't know how to get my delta value because |f(x) -1| = |-x/(1+x)| can't be got into the |x-0| form without worrying about the signs of everything. It's confusing.

Any help?
 
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  • #2
Note that |x-0|=|x| and |-x|=|x|. So,
[tex]
|f(x)-1| = \left| \frac{-x}{x+1} \right| = \left| \frac{x}{x+1} \right|
[/tex]
Now, can you pick your d so that if |x|<d, then |f(x)-1|<p? It helps to allow your d to rely on p.
 
  • #3
n!kofeyn said:
Note that |x-0|=|x| and |-x|=|x|. So,
[tex]
|f(x)-1| = \left| \frac{-x}{x+1} \right| = \left| \frac{x}{x+1} \right|
[/tex]
Now, can you pick your d so that if |x|<d, then |f(x)-1|<p? It helps to allow your d to rely on p.
Well when I get to [tex]|\frac{x}{x+1}|[/tex] and try to find something it's less than I can't get anywhere. So I can't say [tex]|\frac{x}{x+1}| <|x|[/tex] because if x is negative then the denominator is <1. So I don't know how to simplify it to get a value out for delta. Could you give me a bigger hint? Thanks.
 
  • #4
blackcat said:
if x is negative then the denominator is <1
Okay, the denominator is a problem. Can you choose [itex]\delta[/itex] in some way that let's you control what range of values the denominator takes?
 
  • #5
[tex] |\frac{x}{x+1}| = \frac{1}{|1+\frac{1}{x}|}[/tex]

Now use what you know about algebra of limits
 
  • #6
Office_Shredder said:
Now use what you know about algebra of limits
Just to elaborate on this -- while the problem asks you to make an epsilon-delta proof, it does not forbid you from using your knowledge of the algebra of limits to figure out how such a proof should look. You can often directly translate a proof by algebra of limits into an epsilon-delta proof.





By the way, you learned a lot about solving inequalities in precalculus, didn't you? Along with how to solve equations involving absolute values? Have you considered using any of that knowledge?
 
  • #7
Hurkyl said:
Okay, the denominator is a problem. Can you choose [itex]\delta[/itex] in some way that let's you control what range of values the denominator takes?
All I can think of is to make x>0, but then I think that'll go down to taking a limit from the RHS. Are you saying I should take the LHS and RHS limits to get this out?

As for the inequalities knowledge: I guess I have learned quite a bit but I have no idea how else I can approach this problem than the way I'm doing now. Sorry :(
 
  • #8
blackcat said:
As for the inequalities knowledge: I guess I have learned quite a bit but I have no idea how else I can approach this problem than the way I'm doing now. Sorry :(
You have an inequation:
|f(x) - 1| < p​
you have the knowledge to solve that equation for x.
 
  • #9
Ok, so if I solve the inequality, I get |x| < p/(1-p). Is that right?
 
  • #10
just curious, by why work so hard to prove its continuous there?
the right and left handed limits are the same, and the point exists, so its continuous. why does one need epsilon-delta?
 
  • #11
soandos said:
why does one need epsilon-delta?
Because he was specifically asked to do so.

People focus a lot on integrals and differentiation and limits, but understanding how approximation works and how to produce approximations of desired precision and accuracy is one of the basic skills one should be learning in calculus!
 

1. What is the epsilon-delta definition of continuity?

The epsilon-delta definition of continuity is a mathematical concept used to prove that a function is continuous at a specific point. It states that for any given small value of epsilon (ε), there exists a corresponding value of delta (δ) such that if the absolute value of the difference between x and the point of interest is less than delta, then the absolute value of the difference between f(x) and the value of interest is less than epsilon.

2. How is the epsilon-delta definition used to prove continuity?

In order to prove continuity using the epsilon-delta definition, we need to show that for any given small value of epsilon, we can find a corresponding value of delta that satisfies the definition. This involves manipulating the expression for f(x) and setting it equal to epsilon, and then finding a suitable expression for delta in terms of epsilon. If we can show that this value of delta works for all values of epsilon, then we have proven continuity at that point.

3. How can the epsilon-delta definition be applied to prove the continuity of 1/(x+1) at x=0?

To prove that 1/(x+1) is continuous at x=0 using the epsilon-delta definition, we need to show that for any given small value of epsilon, we can find a corresponding value of delta that satisfies the definition. We start by setting the expression for 1/(x+1) equal to epsilon and solving for delta in terms of epsilon. We then show that this value of delta works for all values of epsilon, thus proving continuity at x=0.

4. What are the key steps in proving continuity using the epsilon-delta definition?

The key steps in proving continuity using the epsilon-delta definition are:
1. Set the expression for the function equal to epsilon and solve for delta in terms of epsilon.
2. Show that this value of delta works for all values of epsilon by manipulating the expression for f(x) and the expression for delta.
3. Use the definition to show that the limit of f(x) as x approaches the point of interest is equal to the value of interest.
4. Conclude that the function is continuous at the given point.

5. Are there any limitations to using the epsilon-delta definition to prove continuity?

One limitation to using the epsilon-delta definition is that it can be quite complex and time-consuming to apply, especially for more complicated functions. Additionally, it may not always be possible to find a suitable expression for delta in terms of epsilon, making it difficult to prove continuity using this method. In some cases, alternative methods such as the limit definition of continuity may be more efficient.

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