text: # how to show the following result using continuity defination
  
  • Oct 26th 2012, 11:47 PM
  ujgilani
  how to show the following result using continuity defination
  Using an appropriate definition of continuity, show that the continuity of f (.) at x is equivalent to
  
  lim h->0 sup|t|<h |f (x + t) ─ f (x)| = 0
  • Oct 27th 2012, 12:12 AM
  chiro
  Re: how to show the following result using continuity defination
  Hey ujgilani.
  
  Recall that the limit of lim x->a f(x) = f(a) so what you are doing is instead of considering the epsilon delta definition of |f(x) - l| < e where the limit is l, then this limit l is exactly f(a) = l for continuity. If f(x+t) -> f(x) then that expression should be zero since the limit approaches l = f(x) [Recall: the definition of continuity is that the limit is equal to the value of the function).
  • Oct 27th 2012, 12:12 AM
  johnsomeone
  Re: how to show the following result using continuity defination
  Is this the statement?
  
  $\text{Prop: f is continuous at x } \Longleftrightarrow \lim_{h \to 0+} \left( \underset{|t| < h}{sup} \ \lvert f(x+t) - f(x) \rvert \right) = 0.$
  
  If so, then proof is an $\epsilon - \delta$ proof. To get you started:
  
  $\text{f is continuous at x means } \lim_{y \to x} f(y) = f(x),$
  
  $\text{or equivalently } \lim_{y \to x} (f(y) - f(x)) = 0,$
  
  $\text{or equivalently } \lim_{t \to 0} (f(x+t) - f(x)) = 0,$
  
  $\text{or equivalently } \lim_{t \to 0} |f(x+t) - f(x)| = 0.$
  
  Each of those requires an argument, but is very straightforward. So you can see that the final form there is almost identical to what you're trying to prove, except for that sup (supremum = least upper bound) part.
  
  To prove the proposition, you'll need to prove two things, the $\Rightarrow$ direction, and the $\Leftarrow$ direction.
  
  When you're into the guts of the proof, you'll want to choose (in one direction, though this works for both) the $\delta > 0$ so that
  
  it makes the limit you know $\frac{\epsilon}{2}.$
  
  ----------------------------------------------
  This won't make any sense until you try to prove it yourself (but perhaps then it will be helpful):
  
  If everything in a set is less than $\frac{\epsilon}{2}$, then the supremum of that set will be less than or equal to $\frac{\epsilon}{2}$, which is less than $\epsilon.$
  
  That will be your decisive observation when proving it in one direction.
  
  In the other direction, you'll observe that if the supremeum of a set is less than $\epsilon$ (or you could use $\frac{\epsilon}{2}$ there too),
  
  then any individual one of those things in that set is less than $\epsilon.$
  • Oct 27th 2012, 07:46 AM
  ujgilani
  Re: how to show the following result using continuity defination
  thanks for replies guys .i got it. actually i wanted to post the proof as well but didn't find the way to write post in latex form or the way in which johnsomeone has posted.
  • Oct 27th 2012, 01:51 PM
  ujgilani
  Re: how to show the following result using continuity defination
  hi thanks for your replay, but i'm not getting it, i tried solving it using epsilon-delta method but couldn't complete it another unclear thing is "sup". how to deal with it.
  as i'm new to this stuff so i need little more help in solving it. I'll appreciate if you help me to understand it.
  Quote:
  
  Originally Posted by johnsomeone
  Is this the statement?
  
  $\text{Prop: f is continuous at x } \Longleftrightarrow \lim_{h \to 0+} \left( \underset{|t| < h}{sup} \ \lvert f(x+t) - f(x) \rvert \right) = 0.$
  
  If so, then proof is an $\epsilon - \delta$ proof. To get you started:
  
  $\text{f is continuous at x means } \lim_{y \to x} f(y) = f(x),$
  
  $\text{or equivalently } \lim_{y \to x} (f(y) - f(x)) = 0,$
  
  $\text{or equivalently } \lim_{t \to 0} (f(x+t) - f(x)) = 0,$
  
  $\text{or equivalently } \lim_{t \to 0} |f(x+t) - f(x)| = 0.$
  
  Each of those requires an argument, but is very straightforward. So you can see that the final form there is almost identical to what you're trying to prove, except for that sup (supremum = least upper bound) part.
  
  To prove the proposition, you'll need to prove two things, the $\Rightarrow$ direction, and the $\Leftarrow$ direction.
  
  When you're into the guts of the proof, you'll want to choose (in one direction, though this works for both) the $\delta > 0$ so that
  
  it makes the limit you know $\frac{\epsilon}{2}.$
  
  ----------------------------------------------
  This won't make any sense until you try to prove it yourself (but perhaps then it will be helpful):
  
  If everything in a set is less than $\frac{\epsilon}{2}$, then the supremum of that set will be less than or equal to $\frac{\epsilon}{2}$, which is less than $\epsilon.$
  
  That will be your decisive observation when proving it in one direction.
  
  In the other direction, you'll observe that if the supremeum of a set is less than $\epsilon$ (or you could use $\frac{\epsilon}{2}$ there too),
  
  then any individual one of those things in that set is less than $\epsilon.$
distilabel_metadata.content: Evaluate the user given text extract and determine if it contains a math question and optionally and answer to this question.
  
  Output rules:
  
  1. Briefly describe the document and mention if it contains a question or a question and answer.
  2. Finalize the output with: Label: no-question,question,question-and-answer,multiple-question,multiple-question-and-answer
distilabel_metadata.role: system
distilabel_metadata.content: # how to show the following result using continuity defination
  
  • Oct 26th 2012, 11:47 PM
  ujgilani
  how to show the following result using continuity defination
  Using an appropriate definition of continuity, show that the continuity of f (.) at x is equivalent to
  
  lim h->0 sup|t|<h |f (x + t) ─ f (x)| = 0
  • Oct 27th 2012, 12:12 AM
  chiro
  Re: how to show the following result using continuity defination
  Hey ujgilani.
  
  Recall that the limit of lim x->a f(x) = f(a) so what you are doing is instead of considering the epsilon delta definition of |f(x) - l| < e where the limit is l, then this limit l is exactly f(a) = l for continuity. If f(x+t) -> f(x) then that expression should be zero since the limit approaches l = f(x) [Recall: the definition of continuity is that the limit is equal to the value of the function).
  • Oct 27th 2012, 12:12 AM
  johnsomeone
  Re: how to show the following result using continuity defination
  Is this the statement?
  
  $\text{Prop: f is continuous at x } \Longleftrightarrow \lim_{h \to 0+} \left( \underset{|t| < h}{sup} \ \lvert f(x+t) - f(x) \rvert \right) = 0.$
  
  If so, then proof is an $\epsilon - \delta$ proof. To get you started:
  
  $\text{f is continuous at x means } \lim_{y \to x} f(y) = f(x),$
  
  $\text{or equivalently } \lim_{y \to x} (f(y) - f(x)) = 0,$
  
  $\text{or equivalently } \lim_{t \to 0} (f(x+t) - f(x)) = 0,$
  
  $\text{or equivalently } \lim_{t \to 0} |f(x+t) - f(x)| = 0.$
  
  Each of those requires an argument, but is very straightforward. So you can see that the final form there is almost identical to what you're trying to prove, except for that sup (supremum = least upper bound) part.
  
  To prove the proposition, you'll need to prove two things, the $\Rightarrow$ direction, and the $\Leftarrow$ direction.
  
  When you're into the guts of the proof, you'll want to choose (in one direction, though this works for both) the $\delta > 0$ so that
  
  it makes the limit you know $\frac{\epsilon}{2}.$
  
  ----------------------------------------------
  This won't make any sense until you try to prove it yourself (but perhaps then it will be helpful):
  
  If everything in a set is less than $\frac{\epsilon}{2}$, then the supremum of that set will be less than or equal to $\frac{\epsilon}{2}$, which is less than $\epsilon.$
  
  That will be your decisive observation when proving it in one direction.
  
  In the other direction, you'll observe that if the supremeum of a set is less than $\epsilon$ (or you could use $\frac{\epsilon}{2}$ there too),
  
  then any individual one of those things in that set is less than $\epsilon.$
  • Oct 27th 2012, 07:46 AM
  ujgilani
  Re: how to show the following result using continuity defination
  thanks for replies guys .i got it. actually i wanted to post the proof as well but didn't find the way to write post in latex form or the way in which johnsomeone has posted.
  • Oct 27th 2012, 01:51 PM
  ujgilani
  Re: how to show the following result using continuity defination
  hi thanks for your replay, but i'm not getting it, i tried solving it using epsilon-delta method but couldn't complete it another unclear thing is "sup". how to deal with it.
  as i'm new to this stuff so i need little more help in solving it. I'll appreciate if you help me to understand it.
  Quote:
  
  Originally Posted by johnsomeone
  Is this the statement?
  
  $\text{Prop: f is continuous at x } \Longleftrightarrow \lim_{h \to 0+} \left( \underset{|t| < h}{sup} \ \lvert f(x+t) - f(x) \rvert \right) = 0.$
  
  If so, then proof is an $\epsilon - \delta$ proof. To get you started:
  
  $\text{f is continuous at x means } \lim_{y \to x} f(y) = f(x),$
  
  $\text{or equivalently } \lim_{y \to x} (f(y) - f(x)) = 0,$
  
  $\text{or equivalently } \lim_{t \to 0} (f(x+t) - f(x)) = 0,$
  
  $\text{or equivalently } \lim_{t \to 0} |f(x+t) - f(x)| = 0.$
  
  Each of those requires an argument, but is very straightforward. So you can see that the final form there is almost identical to what you're trying to prove, except for that sup (supremum = least upper bound) part.
  
  To prove the proposition, you'll need to prove two things, the $\Rightarrow$ direction, and the $\Leftarrow$ direction.
  
  When you're into the guts of the proof, you'll want to choose (in one direction, though this works for both) the $\delta > 0$ so that
  
  it makes the limit you know $\frac{\epsilon}{2}.$
  
  ----------------------------------------------
  This won't make any sense until you try to prove it yourself (but perhaps then it will be helpful):
  
  If everything in a set is less than $\frac{\epsilon}{2}$, then the supremum of that set will be less than or equal to $\frac{\epsilon}{2}$, which is less than $\epsilon.$
  
  That will be your decisive observation when proving it in one direction.
  
  In the other direction, you'll observe that if the supremeum of a set is less than $\epsilon$ (or you could use $\frac{\epsilon}{2}$ there too),
  
  then any individual one of those things in that set is less than $\epsilon.$
distilabel_metadata.role: user
distilabel_metadata.raw_output_q_a_detector_0: The document contains a math question and a subsequent discussion that provides an answer and further explanation. The question is about proving the equivalence of the continuity of a function \( f \) at a point \( x \) using a specific limit involving the supremum.
  
  ### Summary:
  - **Question**: The user asks how to show that the continuity of \( f \) at \( x \) is equivalent to the limit:
    \[
    \lim_{h \to 0} \left( \sup_{|t| < h} |f(x + t) - f(x)| \right) = 0.
    \]
  - **Answer**: The responses provide a detailed explanation using the \(\epsilon - \delta\) definition of continuity and the properties of the supremum. The key points are:
    - Continuity at \( x \) means \(\lim_{y \to x} f(y) = f(x)\).
    - This can be rewritten as \(\lim_{t \to 0} |f(x + t) - f(x)| = 0\).
    - The supremum part is handled by noting that if all elements in a set are less than \(\frac{\epsilon}{2}\), then the supremum is