Does this prove the sequence $5+(-1)^n$ does not have a limit? The question is "Consider the sequence $s_n=5+(-1)^n$. Prove that this sequence does not have a limit".
My professor in class proved this by choosing $n_1$ to be even, $n_2$ to be odd, and $n_1,n_2>N$. Then he basically plugged these into and used the definition of a limit to show that the limit does not exist. I have another shorter way I was thinking I could prove this, but I am not sure if it really works.
Here's my idea for a proof:
Let's assume $s_n$ has a limit where $\lim s_n=L$ and $L$ is a real number. Let $\epsilon >0$. Then there exists a $N$ such that $n>N$ implies $|5+(-1)^n-L|<\epsilon$. Then trying to isolate $n$ to find $N$ such that $n>N$ we find $n>\frac{\log(-1)}{\log(\epsilon+L-5)}$. But we see that $\log(-1)$ does not exist, therefore $N$ does not exist such that $n>N$ implies $|5+(-1)^n-L|<\epsilon$. This is a contradiction to our assumption, therefore the opposite must be true. Therefore the limit of the sequence $s_n=5+(-1)^n$ does not exist.
Any information on whether or not this is a valid proof would helpful.
 A: The main problem with your proof is this 

Then trying to isolate $n$ to find $N$ such that $n>N$ we find $n>\frac{\log(-1)}{\log(\epsilon+L-5)}$.

All this tells you is that you've failed to "isolate $n$" appropriately. It does not say that such an $N$ cannot exist. $|5+(-1)^n-L|<\epsilon$ does not imply this. Note, for example, that when $L = 5$ and $\epsilon > 1$, it is true that for every $n,\ |5+(-1)^n-L|<\epsilon$. But even then, your supposed inequality for $n$ makes no sense.
The expression $|5+(-1)^n-L|<\epsilon$ means $$ -\epsilon < 5+(-1)^n-L \text{ and } 5+(-1)^n-L < \epsilon$$
$$L - \epsilon - 5 < (-1)^n \text{ and } (-1)^n < L + \epsilon - 5$$
However, since $\log$ is only defined on positives, you cannot take the logarithm of both sides without first demonstrating that both sides are positive. In this case, they are clearly not positive for all values. So taking the logarithm is not allowable.
What you can see from this is that if this is to hold for all $n > N$ for some N, it must in particular hold for at least one even and one odd value of $n$, and therefore 
$$L - \epsilon - 5 < -1 \text{ and } 1 < L + \epsilon - 5$$ 
$$L - 4 < \epsilon \text{ and } 6 - L < \epsilon$$
Adding the last two inequalities together, we must have $2 < 2\epsilon$, or $1 < \epsilon$. Thus no such $N$ can exist when $\epsilon < 1$.
