# Limits - It's always possible to escape an indeterminate form?

In general, when we are solving limits and get to an indeterminate form($\frac{0}{0}$, $\frac{\infty}{\infty}$, $0^0$,etc.), we use some technique to get rid of the indeterminate form(factorization, conjugate, variable substition, etc.) and then finish solving the limit.

My question is, it's always possible to get rid of indeterminate forms? If it's not, what conclusion should i take about the limit I'm trying to solve?

• I think this is an interesting question. For example, let $X$ denote the set of all elementary real functions vanishing at $+\infty$ but which are non-zero for sufficiently large $x$. Is there an algorithm which can decide whether or not for any given pair $(f,g) \in X \times X$ the limit $$\lim_{x \to +\infty} \frac{f(x)}{g(x)}$$ exists, and if it does, compute it? Sep 3, 2017 at 1:59
• I suspect there are limits which do not exist but give indeterminant forms when one tries to evaluate them. Sep 3, 2017 at 2:23
• For typical problems encountered in calculus courses, the techniques to deal with indeterminate forms are sufficient and one can easily conclude whether the limit exists or not. But this is not the case for every limit problem. For example it took quite sometime for mathematicians to prove that the limit $\lim_{x\to\infty}\dfrac{\pi(x) \log x} {x}$ exists (prime number theorem). Sep 3, 2017 at 3:47
• I'll answer your question "what conclusion should I take..." in the way I answer it in my classroom: Try Something Else. Sep 5, 2017 at 20:04
• @LeeMosher In my opinion, it's always possible to get rid of indeterminate forms. Otherwise, indeterminate forms would need to be contemplated somehow on the formal definition of limits($\delta-\epsilon$). I believe my way of thinking is kind of what you say by try something else. But I'm not sure this is an strong argument. Sep 5, 2017 at 21:53

If you are studying a limit of the form $$\lim_{x\to a}\frac{f(x)}{g(x)},$$ where $f$ and $g$ are nonzero smooth functions with $f(a)=g(a)=0$, then you can get rid of the indeterminate form $\frac00$ if $f$ and $g$ are analytic. In this case you have that $f$ and $g$ can be written as their Taylor's series at $a$ and so, since they are nonzero functions there must be a derivative which is different from zero, say $f^{(n)}(a)$ is the first non-zero derivative for $f$ and $g^{(m)}(a)$ is the first nonzero derivative for $g$. Hence, $$\lim_{x\to a}\frac{f(x)}{g(x)}=\lim_{x\to a}\frac{\frac1{n!}f^{(n)}(a)(x-a)^n}{\frac1{m!}g^{(m)}(a)(x-a)^m}$$ and this limit can be computed. The good news is that most elementary functions in calculus are analytic (polynomias, $\sin x$, $\cos x$, $e^x$, $\log x$) so most of the times indeterminate forms can be solved. However, there are $C^\infty$ functions which are not analytic. As Renart wrote, the typical example is $f(x)=e^{-1/x^2}$. Its Taylor series at $0$ is zero and so if both $f$ and $g$ have this pathology (see Renart's example), then one would would not be able to resolve the indeterminate form with simple tools (Hopital, Taylor's formulas, and the such).

Edit I will try to reply to TMM's criticism. Consider two functions $f$ and $g$ continuous at $a$ with $f(a)=g(a)=0$ and $f(x)\ne 0$, $g(x)\ne 0$ for $x$ near $a$. The formally correct answer to the question "it's always possible to get rid of indeterminate forms" is yes, but only in the sense that either the limit $\lim_{x\to a}\frac{f(x)}{g(x)}$ exists or it does not. Now if the question "is there a general technique to always get rid of indeterminate forms" then the answer is yes for analytic function and no for $C^\infty$ functions. A general technique means a theorem. Hopital's, Taylor's formula, the squeeze theorems are theorems which can be used to compute limits but they are theorems. Now if $f$ and $g$ are $C^\infty$ and not analytic at $a$, with all derivatives $f^{(n)}(a)=g^{(n)}(a)=0$ for all $n$, then there is no general theorem which can determine if the limit exists or not. So yes, if you want, my answer is incomplete but only because there is no complete answer in this case.

• I might be missing something, but this seems right - no idea why it was downvoted (and +1 from me). Sep 10, 2017 at 2:05
• Thanks! It is right. It just follows by writing Taylor's series for both $f$ and $g$ and factoring out the leading terms in both... oh well... Sep 10, 2017 at 2:16
• Thanks for the answer Sep 10, 2017 at 17:39
• The question asks whether it is always possible to get rid of indeterminate forms. This answer only gives a definite answer for analytic functions, and is vague on the remaining cases, only concluding that "simple tools" may not be sufficient. So how does this offer a complete answer? Do there exist indeterminate forms which are not solvable, yes or no? (And my -1 was for you trying to claim a bounty with an incomplete answer. Now the bounty has been awarded and people may stop looking for a complete answer.)
– TMM
Sep 11, 2017 at 8:15
• At least with $\infty/\infty$, analytic functions doesn't mean you'll get rid of the indeterminate form, as exemplified in math.stackexchange.com/a/912694/26369 Sep 12, 2017 at 0:39

There are some examples where l'Hopital does not suffice, and iterating it winds up with the original limit two steps later.

An example of this is the limit of the quotient x / sqrt(x^2 + 1) as x ---> inf. If we apply l'Hopital, we get the limit of sqrt(x^2 + 1) / x, and one more time returns us to the original limit.

Of course, one could argue that IF the limit exists, it therefore must be 1. Or one can argue that for large positive x, numerator and denominator are almost equal, so limit should be 1.

The wikipedia page on indeterminate forms defines them as being restricted to the following seven forms: $$0/0,~ \infty/\infty,~ 0\times\infty,~ \infty-\infty,~ 0^0,~ 1^\infty,~ \infty^0$$ and then goes on to provide a handy table for converting all of these into the form $0/0$ or $\infty/\infty \text{,}$ whose limit can then be found using L'Hôpital's rule

• Sometimes L'Hôpital's rules won't help you. A dumb example is for the limit of $\exp(-1/x^2)/\exp(-1/x^4)$ in $0^+$, you can't use l'Hôpital's rules because it will always give you the indeterminate form $0/0$. This is a "dumb" example because applying the properties of the exponential one can see that the limit is $+\infty$ but it is possible to construct not so dumb similar examples where the l'Hôpital's rules won't work. Sep 6, 2017 at 8:29