Proving the convergence of the sequence defined by $x_1=3$ and $x_{n+1}=\frac{1}{4-x_n}$ Consider the sequence defined by 
$$x_1=3 \quad\text{and}\quad x_{n+1}=\dfrac{1}{4-x_n}$$
I can calculate limit by assuming limit exist and solving quadratic equation, but I first wanted to give existence of limit.
I tried to show given sequence is decreasing and bounded below by 0.
I used derivative test as 
$$f^\prime(x)=\frac{1}{(4-x)^2}$$ 
but form this I am not able to show
Also, I tried to shoe $x_{n+1}-x_n<0$ but that way also I am not succeed.
Please tell me how to approach such problem

You will want to show $x>f(x)$ in order to establish that the sequence is decreasing. So consider $g(x) = x - \dfrac{1}{4-x}.$
Since $g'(x) = 1-\dfrac{1}{(4-x)^2} =\dfrac{(x-5)(x-3)}{(4-x)^2},$ it suggest that your intuition true if $x_n\leq 3$ for $n>1$, which can be easily established by induction.  
