High School Prove Limit Rule: Learn the Constant Concept

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The discussion centers on proving that the limit of a constant function equals the constant itself. Participants emphasize using the definition of a limit, specifically showing that for a constant function f(x) = k, the condition |f(x) - f(a)| < ε holds when |x - a| < δ. One user suggests that since |f(x) - f(a)| equals zero, it satisfies the limit condition for any ε > 0. Others encourage providing a more detailed explanation of how to determine δ in relation to a given ε. The conversation highlights the simplicity of this proof while stressing the importance of clarity in mathematical reasoning.
shihab-kol
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Hello, I would like to begin by saying that this does not fall into any homework or course work for me. It is just my interest.
I need to prove that limit of a constant gives the constant it self. Can some one provide a link? I have exams or I would have searched myself but unfortunately I don't have time. So, I had to post this thread.
 
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shihab-kol said:
Hello, I would like to begin by saying that this does not fall into any homework or course work for me. It is just my interest.
This is one of the easiest of limits to prove, using the definition of a limit to prove it. If f(x) = k, a constant, can you show that ##|f(x) - f(a)| < \epsilon## when ##|x - a| < \delta##?
 
Mark44 said:
This is one of the easiest of limits to prove, using the definition of a limit to prove it. If f(x) = k, a constant, can you show that ##|f(x) - f(a)| < \epsilon## when ##|x - a| < \delta##?
If I take f(a) =k for some interval around x , then |f(x) - f(a)| =0 <E since by definition E >0.
But will this work?
 
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shihab-kol said:
If I take f(a) =k for some interval around x , then |f(x) - f(a)| =0 <E since by definition E >0.
But will this work?
It will work, but you need to say more.
Someone gives you a value of ##\epsilon > 0##. Can you specify a number ##\delta > 0## so that when ##|x - a| < \delta##, then ##|f(x) - f(a)| < \epsilon##?
 
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