- #1

vibha_ganji

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x > 0 for which f(x) is outside N1(A). I don’t get how they generalized the specific statement to all neighborhood of N(0). Thank you!

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In summary, the conversation discusses a proof in Apostol's Calculus (pg. 130) that shows the function 1/(x^2) does not have a limit at 0. It is unclear how the proof concludes that every neighborhood N(0) contains points x > 0 for which f(x) is outside N1(A). The generalization of the specific statement to all neighborhoods of N(0) is based on the existence of at least one point x that maps outside of any 1-neighborhood of the limit A, proving that the limit does not exist. This is a subtle point that requires a thorough understanding of the epsilon-delta definition of a limit.

- #1

vibha_ganji

- 19

- 6

x > 0 for which f(x) is outside N1(A). I don’t get how they generalized the specific statement to all neighborhood of N(0). Thank you!

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- #2

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There is no value ##f(x)## in ##N_1(A)## and you can do this for any ##A##, so ##\lim_{x \to 0^+}f(x) \neq A##. ##A=0## is also impossible, since ##0## isn't a limit point.

- #3

Delta2

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This is a subtle point and requires to have perfect understanding of the exist and for all quantifiers in the epsilon-delta definition of the limit. More specifically if one wants to prove that A is NOT a limit as x->0 then he has to prove that

$$\exists \epsilon>0 : \forall \delta>0 ,\exists x:0<|x-0|<\delta\Rightarrow |f(x)-A|>\epsilon$$

The epsilon is ##\epsilon=1## and the delta is any positive number because for any such delta there exists ##x<min(\delta,\frac{1}{A+2})## for which ##0<|x-0|<\delta## and ##|f(x)-A|>\epsilon=1##. We can choose any ##x<min(\delta,\frac{1}{A+2})## but for the proof we require to exist at least one such x (for every different delta).

So we don't actually have to prove that**all** the points of **every** neighborhood ##N(0)## are such that ##|f(x)-A)|>1##, but just that for every neighborhood## N(0)## there exists at least one point ##x ## such that ##|f(x)-A|>1##.

$$\exists \epsilon>0 : \forall \delta>0 ,\exists x:0<|x-0|<\delta\Rightarrow |f(x)-A|>\epsilon$$

The epsilon is ##\epsilon=1## and the delta is any positive number because for any such delta there exists ##x<min(\delta,\frac{1}{A+2})## for which ##0<|x-0|<\delta## and ##|f(x)-A|>\epsilon=1##. We can choose any ##x<min(\delta,\frac{1}{A+2})## but for the proof we require to exist at least one such x (for every different delta).

So we don't actually have to prove that

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- #4

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This seems like a slightly odd argument to me. The right-hand limit is clearly ##+\infty##, which can be proved quite easily. Then, you must have a general theorem that a limit cannot be both a finite number ##A## and ##+\infty##. That can be done once for all cases.vibha_ganji said:In Apostol’s Calculus (Pg. 130) they are proving that 1/(x^2) does not have a limit at 0. In the proof, I am unable to understand how they conclude from the fact that the value of f(x) when 0 < x < 1/(A+2) is greater than (A+2)^2 which is greater than A+2 that every neighborhood N(0) contains points

x > 0 for which f(x) is outside N1(A). I don’t get how they generalized the specific statement to all neighborhood of N(0). Thank you!

View attachment 288821

For the same effort, you could have a theorem rather than a statement about a single function at a single point.

- #5

pasmith

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vibha_ganji said:In Apostol’s Calculus (Pg. 130) they are proving that 1/(x^2) does not have a limit at 0. In the proof, I am unable to understand how they conclude from the fact that the value of f(x) when 0 < x < 1/(A+2) is greater than (A+2)^2 which is greater than A+2 that every neighborhood N(0) contains points

x > 0 for which f(x) is outside N1(A). I don’t get how they generalized the specific statement to all neighborhood of N(0). Thank you!

View attachment 288821

The intersection of [itex]N(0)[/itex] and [itex](0, 1/(A + 2))[/itex] is [itex](0, x_0)[/itex] where [tex]x_0 = \min\{ \sup N(0), 1/(A + 2)\} > 0[/tex] since both [itex]\sup N(0)[/itex] and [itex]1/(A + 2)[/itex] are strictly positive. This intersection is never empty.

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- #6

WWGD

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Because , as Delta2 suggested, to prove the limit A does not exist, you only need to provide one counterexample: A neighborhood of 0 here that does not map into an ##\epsilon##-neighborhood of the limit, for some value of ##\epsilon##. This is what was provided: a proof for ##\epsilon##=1: terms near 0 will map outside of any 1-neighborhood of any putative limit A.vibha_ganji said:In Apostol’s Calculus (Pg. 130) they are proving that 1/(x^2) does not have a limit at 0. In the proof, I am unable to understand how they conclude from the fact that the value of f(x) when 0 < x < 1/(A+2) is greater than (A+2)^2 which is greater than A+2 that every neighborhood N(0) contains points

x > 0 for which f(x) is outside N1(A). I don’t get how they generalized the specific statement to all neighborhood of N(0). Thank you!

View attachment 288821

In other words: we showed that for a neighborhood of a potential limit A there is no x-axis neighborhood neighborhood mappi ng into it. Any such x-axis 'hood will map outside of the neighborhood (A-1,A+1).

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The limit of a function at a certain point is the value that the function approaches as the input approaches that point.

A function has a limit at a specific point if the value of the function approaches a finite number as the input approaches that point from both sides.

As the input approaches 0 from the positive side, the value of 1/(x^2) approaches positive infinity. And as the input approaches 0 from the negative side, the value of 1/(x^2) approaches negative infinity. Therefore, the function does not approach a finite number and does not have a limit at 0.

Yes, the graph of 1/(x^2) has a vertical asymptote at x=0, indicating that the function approaches positive and negative infinity as x approaches 0 from both sides.

The lack of limit at 0 for 1/(x^2) means that the function is not continuous at that point. This is because the limit of a function at a certain point is a necessary condition for continuity at that point.

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