Find the limit of this sequence $y_0=k$ where $k$ is a constant.
$x_{n+1}=30-\dfrac{y_n}{2}$
$y_{n+1}=30-\dfrac{x_{n+1}}{2}$
Prove that $(x_n, y_n)$ converges to $(20, 20)$ for all values of $k$.
My attempt:
I wrote a computer program and verified this for a few values of $k$. But I don't know how to prove that $x_n$ and $y_n$ converge.
 A: Hint
$$y_{n+1}=30-\frac{30-\frac{y_n}{2}}{2}=\frac{60+y_n}{4}$$
Therefore
$$y_{n+1}-20=\frac{y_n-20}{4}$$
From here you should be able to show that 
$$y_n-20=\frac{y_0-20}{4^n}$$
A: You might be surprised to hear you don't actually have to prove the relation converges before finding what it converges to. 
Assume $\lim_{n \to {\infty}} x_n=a$ and $\lim_{n \to {\infty}} y_n=b$, also assume that the relation converges, this means $x_{n-1}=x_n$. Substitute...
$(1) \quad a=30-\dfrac{b}{2}$
$(2) \quad b=30-\dfrac{a}{2}$
Substitute the right hand side of (2) for the b in (1).
$$a=30-{{30-a/2} \over 2}$$
Solving for a, we get
$$a=20$$
Similarly $b=20$
What does this mean? It means that if the above relation converges, the only value it can converge to is $a=b=20$.
To prove the above has an attractive fixed point, namely at $a=b=20$, we'll take derivatives.
$$y_{n+1}=15+{{y_n} \over 4}$$
$$x_{n+1}=15+{{x_n} \over 4}$$
The derivatives of the above with respect to x and y, in that order are both equal to ${1 \over 4}$. Since the absolute value of the derivative is less than 1 at the fixed points, by the fixed point theorem, all values of $x_0$ and $y_0$ converge to the fixed point.
