I have the sequence $$x_n = \frac{2x_{n-1}^3+7}{3x_{n-1}^2}$$
I know the limit of this sequence is $\sqrt[3]{7}$, and I chose the starting values $x_{0} = 2$, $x_{1} = 1.91666\ldots$
I tried proving by using induction first but failed. How do I prove this sequence does really converge and also its limit is $\sqrt[3]{7}$?
On
Since you know the limit is $\sqrt[3]{7}$, you could introduce a new variable $\epsilon_n = x_n^3 - 7$ and look at the recurrence relation of $\epsilon_n$ instead. We have
$$\epsilon_{n+1} = x_{n+1}^3 - 7 = \frac{(2 x_n^3 + 7)^3}{27 x_n^6} - 7 = \frac{(21 + 2\epsilon_n)^3}{27(7+\epsilon_n)^2} - 7 = \frac{(63+8\epsilon_n)\epsilon_n^2}{27(7+\epsilon_n)^2} $$
Let $\displaystyle\;f(t) = \frac{(63+8t)t^2}{27(7+t)^2}$. Notice for $t \in [-1,1]$,
$$|f(t)| \le \frac{63+8}{27(7-1)^2} |t|^2 \le \frac{71}{972}|t| \quad\implies\quad f(t) \in [-1,1]$$
$f(t)$ is a contraction mapping on $[-1,1]$. Furthermore,
$$\begin{array}{rcl} \epsilon_0 = 1 & \implies & |\epsilon_1| = |f(\epsilon_0)| \le \frac{71}{972}\\ & \implies & |\epsilon_2| = |f(\epsilon_1)| \le \left(\frac{71}{972}\right)^2\\ & \vdots & \\ &\implies & |\epsilon_n| = |f(\epsilon_{n-1})| \le \left(\frac{71}{972}\right)^n\\ & \vdots \end{array} $$ This implies $\lim\limits_{n\to\infty} \epsilon_n = 0$ and hence $\lim\limits_{n\to\infty} x_n = \lim\limits_{n\to\infty}\sqrt[3]{7+\epsilon_n} = \sqrt[3]{7}$.
Hint. Consider the map $$T(x)=\frac{2x^3+7}{3x^2}=\frac{2}{3}x+\frac{7}{3x^2}$$ and the interval $I=[\sqrt[3]{7},2]$. Show that:
i) $T(\bar x)=\bar x$ with $\bar x:=\sqrt[3]{7}$.
ii) Let $T'(x)=\frac{2}{3}\left(1-\frac{7}{x^3}\right)$. Then $0<T'(x)\leq 2/3<1$ for all $x\in I$.
iii) $T(I)\subset I$ ($T$ is increasing by ii), $T(\bar x)=\bar x$ and $T(2)<2$).
Then $T$ is a contraction mapping in $I$.
Hence, by the Banach fixed-point theorem the sequence defined recursively by $x_{n+1}=T(x_n)$ with $x_0\in I$ converges to the unique fixed point $\bar x$.