Request for example of a value of limit of a sequence where sequence does not converge In
Limit of the nested radical $\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$
Timothy Wagner gave a correct answer that was questioned for not having shown that the limit exists in the first place. My question is, are there any examples were a value of limit can be derived although the limit does not exist? Please note I am not questioning that limit must exist before it is exhibited but that, how a value for limit can be exhibited if the limit doesn't exist? Is that not a contradiction? On one hand we have a value for the limit and on the other hand the proof that it can't exist! 
Please give an example were a more general form of convergence does not account for the calculated value. Otherwise it seems as if the value was hinting that the notion of convergence required adjustment and not the calculated value that required justification.
 A: (This should be a comment under Carl's answer.)
@Arjan: One very general way in which to formalize your idea of "possible definitions of convergence" is that of Banach limits. Indeed, a Banach limit is essentially a convergence notion for bounded sequences satisfying a minimal set of reasonable conditions. 
Banach limits do exist---this was proved by Banach himself as a way to show off his Hahn-Banach theorem---and all Banach limits agree with the usual limit on sequences that converge. More interestingly, the set of bounded sequences on which all Banach limits agree is strictly larger than the set of convergence sequences, and it was determined explicitly by [Lorentz, G. G. A contribution to the theory of divergent sequences. Acta Math. 80, (1948). 167--190. MR0027868]
Now, the result of Lorentz implies that not all Banach limits agree on all sequences (in other words, that there is more than one Banach limit) It follows from this that there are divergent sequences which can be assigned at least two different limits under different extensions of the notion of convergence.
A: What he showed was that if a limit $L$ exists, then it has to satisfy the equation $L = \sqrt{7+L}$. This is typical of exercises early in a real analysis class where the only convergence theorem you have available is that bounded monotone sequences converge (this is easy to show given that the reals have the least upper bound property).
So if we have a sequence $(a_n)$ that is defined recursively by $a_1 = x$, $a_{n+1} = f(a_n)$ and is monotone and unbounded, then any fixed point of $f$ is going to be a candidate for a limit, but $(a_n)$ doesn't converge.
A: Yes. Consider the sequence $a_{n+1} = 2a_n + 1$, where $a_0 = 0$. If this sequence converged to some value $L$, then $(a_{n+1})$ would converge to the same value, so we would get $L = 2L + 1$, which gives $L = - 1$. But the sequence actually diverges to infinity.  I use this sort of example when I teach real analysis to drive home the point that we have to check for convergence before trying to compute the value of a limit. 
This method also comes up in with infinite series. Consider the series $1 + -1 + 1 + -1 + \cdots$, which is not convergent. If it were convergent to a value $s$, then we would have $s = 1 + (-1 + 1 + -1 + \cdots)$ so $s = 1 - s$, which leads to $s = 1/2$. So we can compute a "value" even though the series diverges. 
This fact is actually useful in practice when we study generalized notions of convergence. For the moment, define a notion of series convergence to be "nice" if, whenever a series $\sum a_n$ converges, we have $\sum a_n = a_0 + \sum a_{n+1}$ and $\sum -a_n = -\sum a_n$. Then if $1 + -1 + 1 + -1 + \cdots$ converges under any "nice" notion of convergence, its limit will be $1/2$. 
A: The point is often when we try to derive the limit, we invoke algebraic manipulations which would only work if it has already been established that a limit exists (say, by showing the sequence is cauchy). 
