The convergence of a sequence and its limit 
Let the sequence $(x_n)_{n \ge 1}$ be defined by $x_1>0$ and $$x_{n+1}=\frac{x_n}{\sqrt[4]{1+\frac{x_n^8}{n}}}.$$ Prove that  $(x_n)_{n \ge 1}$  is convergent and find  $$\lim_{n\to \infty}x_n.$$

I proved that $x_n$ is convergent, but I don't know how to find the limit. I've tried to create a recurrence between $x_{n+1}$ and $x_{n-1}$, but this doesn't help me. Please help!
 A: Define $y_n = \dfrac{1}{x_n^4}$ for $n \in \mathbb{N}_+$, then$$
y_{n + 1} = \frac{1}{x_{n + 1}^4} = \frac{1}{x_n^4} \left(1 + \frac{x_n^8}{n} \right) = y_n + \frac{1}{n y_n}.
$$
Since $y_1 > 0$, then $\{y_n\}$ is increasing. Suppose that $y_n \not\to +\infty \ (n \to \infty)$, then $l := \lim\limits_{n \to \infty} y_n$ exists and is finite. Now suppose $N \in \mathbb{N}_+$ satisfies that for any $n > N$, $y_n < 2l$. Then for any $n > N$,$$
y_{n + 1} = y_n + \frac{1}{n y_n} > y_n + \frac{1}{2ln},
$$
which implies for any $m \in \mathbb{N}_+$,$$
y_{N + m} = y_{N + 1} + \sum_{k = 1}^{m - 1} (y_{N + k + 1} - y_{N + k}) \geqslant y_{N + 1} + \frac{1}{2l} \sum_{k = 1}^{m - 1} \frac{1}{N + k}.
$$
Let $m \to \infty$, then $l = +\infty$, a contradiction.
Therefore,$$
\lim_{n \to \infty} y_n = +\infty,
$$
which implies$$
\lim_{n \to \infty} x_n = 0.
$$
A: Hint: the occurring operations are continuous on the positive reals, hence the left and right hand sides of the equation converges to the same.
 Can you exclude the case $\lim x_n=0$? 
