Finding x given that $\lim_{n\to\infty } x^{x^{n-1}}=5$ Given a sequence $\{a_n\}^\infty_{n=1}$ defined in the following way: 
$ a_1=x, \ a_2=x^x, \  a_3=x^{x^{x}}, a_4=x^{x^{x^{x}}} ... \ (x&gt;0)$ 
what would be the value of $x$ if  $\lim_{n\to\infty } a_n=5$
It looks to me that you mean tetration. It is commonly written as $^nx$

You mean what is the limit for $lim_{n \to \infty}\,^nx$

That limit only exists for $0 \leq x \leq e^{e^{-1}}$  

If $0 \leq x &lt; e^{-e}$  then the limit is a periodic square wave function, whose max and min values are the solution to $x^{x^c}=c$ (x is the constant, and c the limit). You have 2 limit values, $c_1$ and $c_2$.

If $ e^{-e} \leq x \leq e^{e^{-1}}$  then the limit is the solution to the equation $x^c=c$. The limit c is calculated using the lambert W function.

${}^{\infty}x = x^{x^{\cdot^{\cdot^{\cdot}}}} = c = \frac{\mathrm{W}(-\ln{x})}{-\ln{x}} $

Given that $x=e^{e^{-1}}$ is the maximum value for which you have a limit, there is no limit for $^nx=5$.

But there is a limit for $lim_{n \to \infty} \,^{-n}x=5$, if $^0x$ is defined as $^0x&gt;lim_{n \to \infty} \,^{n}x$. In such case, $x\approx 1,3797296901083$

That's because there is some arbitrariness in defining $^0x$. It is commonly defined as $^0x=1$, but that's just a (sensible) choice.

Yet $^nx$ is not $x^{x^{n-1}}$ (as you wrote in the title). The solution to $\lim_{n\to\infty } x^{x^{n-1}}=z$ is z because for the solution to exist, it should be $z^z=z$. But z=5 does not meet the requirements.
Let 

$$\zeta = x^{x^{x^{...}}}$$

We know that 

$$\zeta = 5$$

Since:

$$\zeta = x^{x^{x^{...}}}$$

We can say that: 

$$\zeta = x^{\zeta}$$

Thus, you can solve for x using $5 = x^5$.