Proof of 1/(x^2) Not Having Limit at 0

[vibha_ganji]

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!

 

[fresh_42]

The proof is correct. Draw a picture:

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.

 

[Delta2]

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 $|f(x)-A|>1$.

 

[PeroK]

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

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.

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

 

[pasmith]

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 N(0) and (0, 1/(A + 2)) is (0, x_0) where x_0 = \min\{ \sup N(0), 1/(A + 2)\} &gt; 0 since both \sup N(0) and 1/(A + 2) are strictly positive. This intersection is never empty.

 

[WWGD]

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

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.

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).