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?
 A: 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.
A: 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.
A: 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
