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Suppose one has the limit of a completely non-continuous when $x<0$ such as $$\lim _{x\to-\infty}\left|{x}^{\frac{1}{2x}}\right|$$ Which is undefined at any negative fraction with an odd denominator (such as $-2$, $-2/3$, $-3/5$).

If you treat this function as a sequence at any integer it is undefined. However, by using integers of a fraction of even denominator such as $-1/2$ then $-1/4$.

$$ \begin{array}{c|lcr} x & \text{y} \\ \hline -\frac{1}{2} & 2.. \\ -\frac{3}{2} & .87358.. \\ -\frac{5}{2} & .83255.. \\ -\frac{7}{2} & .83613..\\ -\frac{10001}{2} & .99915..\\ \end{array} $$

I did find an identity that $\lim _{x\to-\infty}\left|{x}^{\frac{1}{2x}}\right|=\lim_{x\to-\infty}\left|{\left(-x\right)}^{\frac{1}{2x}}\right|$ and you could take l'hospitals rule.

$$\lim_{x\to-\infty}\left|\pm{\left(-x\right)}^{\frac{1}{2x}}\right|$$ $$\lim_{x\to-\infty}\left|{\pm}e^{\frac{\ln{\left(-x\right)}}{{{2x}}}}\right|$$ $$\lim_{x\to-\infty}\left|{\pm}e^{\frac{\frac{1}{x}}{{{2}}}}\right|$$ $$\lim_{x\to-\infty}\left|{\pm}e^{0}\right|$$ $$\lim_{x\to-\infty}\left|{\pm}e^{0}\right|=1$$

Has there been any analysis done in these areas that I could find. Is this how to solve these kinds of limits?

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  • $\begingroup$ Is it consistent with what you get if you consider f has values in the complex? $\endgroup$ – Jean-François Gagnon Oct 15 '15 at 13:42
  • $\begingroup$ No it is consistent only with real value input that give real number output. However the real values outputs approach a limit but not at integers. $\endgroup$ – Arbuja Oct 15 '15 at 13:45
  • $\begingroup$ If you cannot answer this question, then what are the exact requirements for a limit to be possible? $\endgroup$ – Arbuja Oct 15 '15 at 15:09
  • $\begingroup$ If the limite does not exist for a sub sequence then it does not exist. $\endgroup$ – Jean-François Gagnon Oct 15 '15 at 15:32
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    $\begingroup$ Your function $x \mapsto \left\lvert x^{\frac{1}{2x}}\right\rvert$ is actually continuous, because continuity is determined at and only at points in the domain of the function. The correct thing to say instead is that the domain of definition of the function is extremely disconnected. $\endgroup$ – epimorphic Oct 25 '15 at 18:57
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Regarding the limit : If the limit does not exist for a subsequence then it does not exist.

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  • $\begingroup$ But why is this the case? Like I said any the limit exists with $-\frac{1}{2}$, $-\frac{3}{2}$, $-\frac{5}{2}$., $\frac{-7}{2}$... $\endgroup$ – Arbuja Oct 15 '15 at 17:43
  • $\begingroup$ Nevermind, you're leaving it to me to decide whether ithe limit exists or not. There is a subsequence so it does. Thank You! $\endgroup$ – Arbuja Oct 15 '15 at 17:56
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    $\begingroup$ No. Please read what I wrote more carefully. $\endgroup$ – Jean-François Gagnon Oct 15 '15 at 19:00
  • $\begingroup$ I meant to say "there is a limit for the subsequence". For example if you take instead of $S=\lim_{x\to\infty}={a}_{k}$ we take $S=\lim_{x\to\infty}{a}_{k-\frac{1}{2}}=1$ $\endgroup$ – Arbuja Oct 15 '15 at 19:11
  • $\begingroup$ I meant negative infinity for the above post. $\endgroup$ – Arbuja Oct 15 '15 at 19:28
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The limit exists and is indeed $1$. Here's a standard general definition of the limit of a function on a subset of the reals:

Let $A$ be a subset of the extended real line $[-\infty, \infty]$, let $f \colon A \to [-\infty, \infty]$ be a function, and let $p$ be a limit point of $A$ in $[-\infty, \infty]$. We say that $l \in [-\infty, \infty]$ is the limit of $f$ at $p$ if for each open interval $V \subset [-\infty, \infty]$ containing $l$, there is an open interval $U \subset [-\infty, \infty]$ containing $p$ such that $f\bigl((U \cap A) \setminus \{p\}\bigr) \subset V$.

The relevant sections in the Wikipedia article "Limit of a function" are "Functions on topological spaces" and "Limits involving infinity".

Basically, we can talk about whether the limit of $f$ exists at $-\infty$ as long as $f$ is defined at arbitrarily large negative numbers (i.e. the domain of $f$ contains a sequence tending to $-\infty$). We then say that $\lim_{x \to -\infty} f(x) = l \in \mathbb R$ if for each $\epsilon > 0$, there exists some $b \in \mathbb R$ such that $\lvert f(x) - l\rvert < \epsilon$ for all $x < b$ in the domain of $f$. We simply ignore the holes where $f$ isn't defined.

In the present case, you found correctly that the function $x \mapsto \left\lvert x^{\frac{1}{2x}}\right\rvert$ is definable on sets containing arbitrarily large negative members. Moreover, you found that wherever it is defined on the negatives, it coincides with the function $x \mapsto \left\lvert (-x)^{\frac{1}{2x}}\right\rvert$. Since the latter function has limit $1$ as $x \to -\infty$, so does the former.

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  • $\begingroup$ If this is the case is it still possible to take the derivative and place a tangent line at a specific point in the negative domain that is defined? $\endgroup$ – Arbuja Oct 29 '15 at 12:36
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    $\begingroup$ Apologies for the lateness in reply. Unlike continuity, one-dimensional derivatives are generally defined only on open subsets of $\mathbb R$. On the other hand, the only reason I can think of for the difference in convention is that you can't define derivatives at isolated points of the domain. In this case, the domain of your function is dense in the negatives, so it does seem meaningful to me to talk about local behavior. So... I don't see why not. $\endgroup$ – epimorphic Oct 30 '15 at 20:24

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