Let \rightarrow \Re" alt="f \rightarrow \Re" /> be continuous at and suppose that . Prove that there exists an and a neighborhood U of c such that for all .
Let \rightarrow \Re" alt="f \rightarrow \Re" /> be continuous at and suppose that . Prove that there exists an and a neighborhood U of c such that for all .
Let since f is continous there exists a such that when
Now working with the 2nd statement
Now add f(c) to all sides of the inequalilty to get
I understand your proof, but where does the come in? And I can see what you did with the neighborhood but the question confused me because it says an I don't understand the intersection and\or what means. Why must be in the domain ?
I understand your proof, but where does the come in? And I can see what you did with the neighborhood but the question confused me because it says an I don't understand the intersection and\or what means. Why must be in the domain ?
The a is the value of epsilon that I used. Note it is not unique I could have used any fractional part of f(c). The interscetion is to make sure that my neighborhood stays inside the domain of the function. Otherwise it doesn't make sense to talk about function values.