That is indeed trivial. In fact so trivial that I doubted if I would even respond.
Notice that for (real) b >> (z+1) exp(1/e) and (real) z > 0 :
b^^z > z.
Hence lim b^^z >= lim z.
If you are not convinced of b^^z > z Consider that
1) b^z > z for z and b sufficiently large.
2) b^^0 (=1) > 0
3) the derivative of b^^z > 1 whereas the derivative of z is 1.
By induction that is clear.
However I must add that I did assume a nice tetration here.
In other words I assumed for c>>b that c^^z >> b^^z and that c^^z and b^^z are continu.
It always matters what type of tetration you speak of.
However they share similar properties e.g. all infinitely differentiable real solutions to exp^[1/2](x) agree on their values infinitely often.
Notice it is nicer if a half iterate is strictly rising , otherwise when it is both rising and decending taking a half derivitive of that is troublesome.
On a piece of paper that might make more sense to you.
Perhaps usefull to note is that in your question your value of b does not depend on 1/z. That is important since b^^0 = 1 , so if z goes to 0 much faster than b goes to oo ...
regards
tommy1729
Notice that for (real) b >> (z+1) exp(1/e) and (real) z > 0 :
b^^z > z.
Hence lim b^^z >= lim z.
If you are not convinced of b^^z > z Consider that
1) b^z > z for z and b sufficiently large.
2) b^^0 (=1) > 0
3) the derivative of b^^z > 1 whereas the derivative of z is 1.
By induction that is clear.
However I must add that I did assume a nice tetration here.
In other words I assumed for c>>b that c^^z >> b^^z and that c^^z and b^^z are continu.
It always matters what type of tetration you speak of.
However they share similar properties e.g. all infinitely differentiable real solutions to exp^[1/2](x) agree on their values infinitely often.
Notice it is nicer if a half iterate is strictly rising , otherwise when it is both rising and decending taking a half derivitive of that is troublesome.
On a piece of paper that might make more sense to you.
Perhaps usefull to note is that in your question your value of b does not depend on 1/z. That is important since b^^0 = 1 , so if z goes to 0 much faster than b goes to oo ...
regards
tommy1729
