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Hi I'm sorry for bothering you a lot, but i'm practising for my last test When ever i try to do something it's inverting to NaN situation
Can any one help me
Thanks a lot
Note : Obviously It's not 2sided limit (0+)(0-)
With L'Hôpitale I've found that

right side limit is +∞

left side limit is -∞

$$ \bbox[] { \lim_{x\to0} \left( \frac{e^x-x^x}{x^2} \right) \ } $$

AbdulQader Qassab
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  • Clearly the limit does not exist, because the left limit , $x\to 0^{-}$, is not well defined. – FormerMath Oct 10 '16 at 21:54
  • The function isn't defined for $;x<0;$ and thus the limit doesn't exist – DonAntonio Oct 10 '16 at 21:54
  • @DonAntonio infact, I want it 2sided limit – AbdulQader Qassab Oct 10 '16 at 21:55
  • @Luis I've edited the post , thanks for your hint – AbdulQader Qassab Oct 10 '16 at 21:57
  • @user11618 Exactly, and thus the answer is: the limit doesn't exist when $;x\to0;$ as the function isn't even defined for $;x<0;$ . The only thing possible to do here is to take the one-sided limit $;x\to0^+;$ . – DonAntonio Oct 10 '16 at 21:57
  • @DonAntonio: I don't agree. The limit is defined based on the points in the domain of the function. If the function is only defined for $x>0$, the limit is implicitly $0^+$. –  Oct 10 '16 at 22:02
  • In wolfram alpha the complex parts seem to diverge for $x\to 0^- $ – Emil Oct 10 '16 at 22:05
  • @YvesDaoust I don't agree with you, but it doesn't matter: the OP already made it clear he wants the two sided limit , and that thing doesn't exist in this case. – DonAntonio Oct 10 '16 at 22:06
  • @DonAntonio: it suffices that the point where you want to evaluate the limit be adherent to the domain (and the $\epsilon/\delta$ rule to hold) for the limit to exist. –  Oct 10 '16 at 22:14
  • @YvesDaoust Again, I beg to difer. In basic calculus we (I, in fact) teach, and I think most basic text books agree (but I'd love to read other opinions on this) that when the limit $;x\to x_0;$ is considered, the function must be defined in a complete, two-sided, neghborhood of the point $;x_0;$ to even begin considering the existence of the limit. That's why, among other possible reasons, is that the idea of one-sided limit exists and is defined. – DonAntonio Oct 10 '16 at 22:21
  • DonAntonio is correct! Usually when one wants to find a one sided limit - it is explicity specified that it needs to be found. General "What is the limit of ... " questions, need both sides to approach the same limit in order for it to exist.

    For example $y=\frac{1}{x}$ clearly does not have a limit about $(0,0)$ as it is different depending on what direction you approach it.

     – Hugh Entwistle Oct 10 '16 at 22:26
  • @DonAntonio I beg to differ (and I concur with Yves Daoust). Under my convention, when you write $\lim_{x\to x_0^+}f(x)$ you're just considering the function $f$ restricted to $D(f)\cap(x_0,\infty)$ and taking the limit of this function at $x_0$ (where $D(f)$ means the domain of $f$). No need for separate definitions: it's always the same. – egreg Oct 10 '16 at 22:29
  • @HughEntwistle This case is very different: the function is not defined for $x<0$ to begin with. – egreg Oct 10 '16 at 22:29
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    @egreg11 Yes it is a different "case" but it is still an example of a limit that is different when approaching it along two different sides. As others have said, if the function isn't properly defined at $x<0$ then you cannot approach it from $x<0$ ? Thus $\lim_{x\rightarrow0^+}\neq \lim_{x\rightarrow0^-}$ which means the two sided limit does not exist – Hugh Entwistle Oct 10 '16 at 22:34
  • @egreg I invite both you and Yves to show me some text, link or whatever that agrees with you, and I'm not saying this patronizing. Honestly. It really interests me to learn whether conventions have, perhaps, changed so much that I haven't yet payed attention to them. I still remember an exam in undergraduate mathematics that my daughter took and in which there was a question more or less similar to this one, and the expected answer was "the limit cannot exist as there doesn't exist (a whole) niehgborhood of $;x_0;$ where the function is defined", – DonAntonio Oct 10 '16 at 22:35
  • Cont. which is precisle what the condition $;|x-x_0|<\delta;$ means: not one sided, but the whole neighborhood. – DonAntonio Oct 10 '16 at 22:35
  • @HughEntwistle It's a matter of conventions; one shouldn't presume that their own are the only possible ones. – egreg Oct 10 '16 at 22:36
  • @egreg I agree with your last comment, and that's why I did the invitation in my last comment. Thanks – DonAntonio Oct 10 '16 at 22:36
  • @egreg indeed, but I feel that limits need to exist in both sides, as per the limit definition that DonAntonio referred to above. But I will be watching this question closely to see if I need to revise my limits! – Hugh Entwistle Oct 10 '16 at 22:37
  • @DonAntonio $0<|x-x_0|<\delta$ and $x\in D(f)$: this is too often neglected. This is sufficient for a complete and uniform theory of limits. Of course $x_0$ must be an accumulation point of $D(f)$. – egreg Oct 10 '16 at 22:38
  • @egreg My invitation stands. I just began checking some of my text books but these ones could be outdated and now it is more widely used as you said. And in fact, $;x_0\in D(f);$ would mean an interior point, either by condition or else by requiring, as many authors do, that the domain of definition of a function be an open set (many times, also "c\onnected" is required). In the reals this would solve all this mess, of course... – DonAntonio Oct 10 '16 at 22:44
  • Try to use $$\lim_{x\to0} \left( \frac{e^x-x^x}{x^2} \right)=\lim_{x\to0} \left( \frac{e^{x/2}-x^{x/2}}{(\frac x2)^2} \right)(\frac{e^{x/2}+x^{x/2}}4)$$ – Pentapolis Oct 10 '16 at 23:12
  • @DonAntonio: As I said, $x_0$ needs to be an adherence point, which may or not belong to the domain. This makes no difference as the evaluations are at $x\ne x_0$. –  Oct 11 '16 at 07:15
  • @DonAntonio: Wikipedia (from Rudin) does mention the restriction to the domain https://en.wikipedia.org/wiki/(%CE%B5,_%CE%B4)-definition_of_limit#Precise_statement –  Oct 11 '16 at 07:18
  • @YvesDaoust Thanks. I suppose some others do, yet the usage seems to be heterogeneous. For example, Swokowski's "Calculus with Anal. Geom." defines limit as $;x\to a;$ for a function defined on an open interval containing $;a;$, which btw is the point for the well-known concept of punctured neighborhood. Also Adams' "Calculus" mentions this more clearly, when giving an informal definition of limit before the formal one. Stewart' "Calculus: early transcendentals" does even more: he specifies that values of $;x;$ must be taken "sufficiently close to $;a;$ (both sides) but not..." – DonAntonio Oct 11 '16 at 11:14
  • Cont. ...equal to $;a;$" . Also the well known Thomas-Finney's "Calculus with Analytic Geometry" requires in its formal definition of $;f;$ that it must be defined in an open interval "about $;x_0;$ " . I think this may be just a matter of usage, but imo the definition of one-sided limits makes it necessary to draw a line between "limit" (both sides) and "from the right or from the left" limits. – DonAntonio Oct 11 '16 at 11:19

2 Answers2

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Take $x\to 0^+$ and rewrite $x^x=e^{x\ln x}$. Thus $$ \frac{e^x-x^x}{x^2}=\frac{1}{x}\cdot \bigg[\frac{e^x-1}{x}-\frac{e^{x\ln x}-1}{x\ln x}\cdot\ln x\bigg]\to +\infty\cdot[1-1\cdot (-\infty)]=+\infty. $$

A.Γ.
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HINT:

For $x>0$, we have $x^x=1+x\log(x)+O(x\log(x))^2$.

And $\lim_{x\to 0^+}\frac{1-\log(x)}{x}=\infty$

Mark Viola
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  • Nice, but observe that using your idea (which I'm not sure the OP can use it as it is with derivatives...), the limit is $;+\infty;$ , since $;\log x\to -\infty;$ when $;x\to0^+;$ . – DonAntonio Oct 10 '16 at 22:11
  • But how can I use this, It's not useful because there's going to be a new NaN situation(+∞-∞) – AbdulQader Qassab Oct 10 '16 at 22:14
  • @DonAntonio Thank you for catching typo. I posted this way forward cognizant of the fact that the Taylor series is independent of L'Hospital's Rule. – Mark Viola Oct 10 '16 at 22:16
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    @user11618 Actually, it renders the problem of the form $\frac\infty0$, which is not indeterminate. – Mark Viola Oct 10 '16 at 22:17