<s>[INST] Limit of $x_n/n$ for sequences of the form $x_{n+1}=x_n+1/x_n^p$ 
*

*Given $x_1 = 1, x_{n+1} = x_n + \frac{1}{x_n} (n\ge1)$, Prove whether the limit as follow exist or not. If so, find it
$$\lim_{n\to\infty}\frac{x_n}{n}$$

*Given $x_1 = 1, x_{n+1} = x_n + \frac{1}{\sqrt{x_n}} (n\ge1)$, Prove whether the limit as follow exist or not. If so, find it
$$\lim_{n\to\infty}\frac{x_n}{n}$$

In both cases, $x_n$ is increasing, so I tried to get an upper bound on $x_n$ (possibly depending on $n$) to apply a squeeze theorem; but failed.
 [/INST] By the inspiration from @kedrigern, I solved the second one. Hope somebody can find a simple way to solve it:
Let $y_n = x_n^2$,then:  
\begin{align*}
\ y_{n+1} &= x_{n+1}^2 = x_n + \frac{1}{\sqrt{x_n}}= x_n^2 + 2\sqrt{x_n} + \frac{1}{x_n} 
\\&= y_n + 2\sqrt{x_n} + \frac{1}{x_n}
\\&=y_{n-1} + 2(\sqrt{x_n} + \sqrt{x_{n-1}}) +(\frac{1}{x_n}+\frac{1}{x_{n-1}})
\\&\vdots
\\&= y_1 + 2\sum_{i=1}^n\sqrt{x_i} + \sum_{i=1}^n\frac{1}{x_i}
\end{align*}
(1)Since $1\le x_n \le n$. then:
\begin{align*}
\ y_{n+1} &=  1 + 2\sum_{i=1}^n\sqrt{x_i} + \sum_{i=1}^n\frac{1}{x_i}
\\&\ge 1+2\sum_{i=1}^n1+\sum_{i=1}^n\frac{1}{x_i} 
\\&=1+2n+\sum_{i=1}^n\frac{1}{x_i}
\\&\ge n 
\end{align*} 
Hence：
$$ \frac{1}{x_{n+1}} \le \frac{1}{\sqrt{n}}~~That~~is: \frac{1}{x_{n}} \le \frac{1}{\sqrt{n-1}}(n\ge 2)$$
(2)Since $1\le x_n \le n$,then：
$$\sqrt{x_n} \le \sqrt{n}$$  
(3)The upper bound of $y_{n+1}$：
\begin{align*}
\ 0\le y_{n+1}&= 1+2\sum_{i=1}^n\sqrt{x_i}+\sum_{i=1}^n\frac{1}{x_i} 
\\&\le 1+2\sum_{i=1}^n\sqrt{i}+\sum_{i=1}^n\frac{1}{x_i}~~~~~~~~~~~~~~~~~~~~~\ldots\ldots from~(2)
\\&= 1+2\sum_{i=1}^n\sqrt{i}+(1+\sum_{i=2}^n\frac{1}{x_i}) 
\\&\le 1+2\sum_{i=1}^n\sqrt{i}+(1+\sum_{i=2}^n\frac{1}{\sqrt{i-1}})~~~\ldots\ldots from~(1)
\\&= 1+2(\sqrt{n}+\sum_{i=1}^{n-1}\sqrt{i})+(2+\sum_{i=3}^n\frac{1}{\sqrt{i-1}})  
\\&\le 1+2(\sqrt{n} + \int_1^n\sqrt{x}dx)+(2+\int_2^n\frac{1}{\sqrt{x-1}}dx)
\\&=2\sqrt{n}+2\sqrt{n-1}+\frac{4n^\frac{3}{2}-1}{3} 
\end{align*}
Simplify,we have：
$$0\le y_{n+1} \le 2\sqrt{n}+2\sqrt{n-1}+\frac{4n^\frac{3}{2}-1}{3} $$ 
That is：$$0 \le \frac{x_{n+1}}{n} \le \frac{\sqrt{2\sqrt{n}+2\sqrt{n-1}+\frac{4n^\frac{3}{2}-1}{3}}}{n}$$ 
By the squeeze theorem：
$$ \lim_{n \rightarrow \infty}\frac{x_n}{n}=0 $$
</s>