I'm confused about a point that my book on real analysis is making about the difference between(adsbygoogle = window.adsbygoogle || []).push({}); the definition of a function being continuosat at point and thedefinition of a function having a limitL at a point.

My rough understanding of the matter is that in order for a function to be continuous at a point the function must have a limit L at the point (from both sides) and the values of the function at that point must be L. However it is possible to have a limit at a point, even if the value of that function at the point is not the same as the limit.

From the book:

DEFINITION: Suppose E is a subset of R and [tex]f: E \rightarrow R[/tex]. If [tex]x_0 \in E[/tex], then f is continuous at [tex]x_0[/tex] iff for each [tex]\epsilon > 0[/tex], there is a [tex]\delta > 0[/tex] such that if

[tex]|x-x_0|< \delta[/tex], with [tex]x \in E[/tex],

then

[tex]|f(x)-f(x0)|< \epsilon[/tex].

Compare this with the definition of the limit of a function at a point [tex]x_0[/tex] . First of all, for continuity at [tex]x_0[/tex], the number must belong to E but it need not be an accumulation point of E. (Is this saying that f(x) = L? If not what is it saying?) Indeed, if [tex]f: E \rightarrow R[/tex] with [tex]x_0 \in E[/tex] and [tex]x_0[/tex] not an accumulation point of E, then there is a [tex]\delta > 0[/tex] such that if [tex]|x - x_0 | < \delta[/tex] and [tex]x \in E[/tex], then [tex]x=x_0[/tex] .

That last bit has me lost. Why would [tex]x = x_0[/tex] if [tex]x_0[/tex] is not an accumulation point of E?

I think that [tex]x_0[/tex] , not being an accumulation point means that there is at least one neighborhood of [tex]x_0[/tex] that contains a finite number, possibly zero vales of f(x) when x is in a delta neighborhood of [tex]x_0[/tex]. But how does this force [tex]x = x_0[/tex]?

The book defines accumulation points interms of series:

A real number A is an accumulation point of S iff every neighborhood of A contains infinitely many points of S.

So, not an accumulation point would mean that every neighborhood of A contains finite points of S, or zero ponits of S... ?

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# Homework Help: Difference between a function being continuos at at point and havinga limit there

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