(Non)Existence of limits

When we say that a limit of a function does not exist in $\mathbb{R}$ (or some metric space) does it make sense to say that it might exist somewhere else? [I am trying to think along lines of existence of imaginary roots]

If yes. Then give examples especially regarding $\mathbb{R}$.

Yes for example a sequence of rational numbers might converge to $\sqrt{2}$ which is not rational. . So the sequence does not converge in $\Bbb Q$ but does converge in $\Bbb R$.

If your sequence is in $\Bbb R$ though, then since $\Bbb R$ is complete, if the sequence is Cauchy then it does converge somewhere. So if it doesn't converge in $\Bbb R$ it won't converge anywhere bigger because it won't be Cauchy.

Another example is $$\lim_{x\to 0}\frac1{x^2}$$ which does not exist in $\mathbb R$, but does exist in the extended reals $[-\infty,\infty]$.

It also exists in the extended complex numbers (Riemann sphere, roughly $\hat{\mathbb C}=\mathbb C\cup\{\infty\}$).

Note that $$\lim_{x\to 0}\frac1{x}$$ does not exist in the reals or the extended reals, but does exist in the Riemann sphere. That's because there is only one infinity in the Riemann sphere, but not so in the extended reals.

• Explain please. I was demanding an answer along these lines. – Non-Being Feb 14 '16 at 17:56
• But R being a complete m. s. Combined with Gregory grant's reasoning makes sense – Non-Being Feb 14 '16 at 17:57
• Completeness depends on the metric. The reals are homeomorphic to $(0,1)$, but the latter is not complete in the usual metric. So it is not a topological property. – MPW Feb 14 '16 at 18:05
• The problem is the extended reals are not a metric space. So if by "somewhere else" you mean a metric space then it can't ever happen if the smaller space is complete. BUT if by "somewhere else" you just mean any topological space, then yes the extended reals is another example. But you have to modify your definition of convergence to the general topological space - in other words you cannot talk about "distance" you have to just work with open sets. – Gregory Grant Feb 14 '16 at 18:08
• I would extend reals with slightly different way. Use 1-point compactification to get a circle, so both limits(that are given as examples) do exist, basically because there is single infinity. – Mihail Feb 14 '16 at 22:18

For example the sequence $a_n = \bigg(1+\frac{1}{n}\bigg)^n$ has all $a_n \in \mathbb{Q}$ but $$\lim_{n\rightarrow \infty} \bigg(1+\frac{1}{n}\bigg)^n =e \notin \mathbb{Q}$$

If however, $a_n \in \mathbb{R}$ then $(\lim_{n\rightarrow \infty} a_n) \in \mathbb{R}$ (if it exists) since $\mathbb{R}$is complete.

I am not sure if this might be helpful, but I bumped into the following example while answering this question Squeeze theorem on expression with minus. There, help was required for the following limit $$\lim_{x \rightarrow \infty }(-6e^{11x} + 9\sin(x) + 3e^{8x} )^{8/4x}\ .$$

The function $f(x)=(-6e^{11x} + 9\sin(x) + 3e^{8x} )^{8/4x}$ is not defined for $x$ real and large, because $h(x)=-6 e^{11x}+9\sin(x)+3 e^{8x}$ becomes badly negative there, so the limit is somehow ill-defined in $\mathbb{R}$. However, one might define a function $g:\mathbb{R}\to\mathbb{C}$ $$g(x)=\exp\left(\frac{8}{4x}\mathrm{Log}(-6e^{11x} + 9\sin(x) + 3e^{8x} )\right)\ ,$$ with $\mathrm{Log}$ the principal branch of the complex logarithm https://en.wikipedia.org/wiki/Complex_logarithm. The function $g(x)$ is 'morally' equivalent to $f(x)$ [and it coincides with $f(x)$ for all $x$ such that $h(x)>0$], but is perfectly well defined for all $x\in\mathbb{R}$. So, we might ask: what is $\lim_{x\to\infty}g(x)$ [the limit of a complex-valued function of a real argument]?

For $z=x+\mathrm{i}y$, using $\mathrm{Log}(z)=\ln |z|+\mathrm{i}\ \mathrm{atan2}(y,x)$, where $\mathrm{atan2}$ is defined in https://en.wikipedia.org/wiki/Atan2, one obtains $$g(x)=\exp\left[\frac{8}{4x}\left(\ln(6e^{11x} - 9\sin(x) - 3e^{8x} )+\mathrm{i}\pi\right)\right]\ ,$$ so that the imaginary part vanishes in the limit $x\to\infty$ and one obtains $$\lim_{x\to\infty} g(x)=e^{22}\ ,$$ the result given by Mathematica. Note, however, that Mathematica gives the same result for the limit $\lim_{x\to\infty}f(x)$, even though (as I remarked above) this is not defined for $x\in\mathbb{R}$. In summary, in this example we could give a meaning to an ill-defined limit of a real function by moving 'somewhere else' (i.e. allowing the function to take imaginary values for real arguments). Not sure if this was what the OP wanted to know, honestly.

While it may be tempting to "enlarge" $\mathbb{Q}$ to $\mathbb{R}$, it is tautological to say that a limit exists there. At least according to the standard definition where real numbers are equivalence classes of Cauchy sequences. Really, $a_n : \mathbb{N} \rightarrow \mathbb{Q}$ and $a_n : \mathbb{N} \rightarrow \mathbb{R}$ are different functions. All you can say is that the limit of the second one exists and the limit of the first one does not.

Of course we rarely use "exists" this way in common speech, but its definition depends on things we have defined in the problem. Otherwise we'd be able to say that $\lim_{x \rightarrow 0} \log(x)$ exists. It's not a complex number, but if we just dream up some "$\log(0)$" from the "set of all sets", the limit could be said to exist in $\mathbb{C} \cup \{\log(0)\}$ which is just silly.