Fix point interval tends to the fix point A point $x_0$ is called a fixpoint of a function $f$ if $f(x_0) = x_0$. Suppose $I = (x_0 −c, x_0 +c)$ is an open interval containing a fixpoint $x_0$ of a continuously differentiable function f and suppose there exists a rational number $r$, $0<r<1$, such that $|f′(x)| ≤ r < 1$ for all $x ∈ I$. Let a be a point in I and define a sequence $(a_n)$ by: $a_0 =a$ and $a_{n+1} =f(a_n)$ for $n≥0$. 
Prove that $\lim_{n-> \infty} a_n = x_0$ 
I was told that using the mean value theorem might be a good idea, such that $\frac{a_{n+1} - x_0}{a_n - x_0}  =  \frac{f(a_n) - f(x_0)}{a_n - x_0}$  but I can't figure out what to do with this information.  
 A: Though I don't see it written anywhere, I'm assuming that $a\in I$ is some chosen starting point. How about we prove by induction that 


*

*$a_n\in I$, and 

*$|a_n - x_0|\leq r^n|a-x_0|$


for all $n\geq 0$. Since $r < 1$, condition (2.) implies that $|a_n - x_0|\to 0$ as $n\to \infty$, which is what you want.
The base case of $n = 0$ is clear. For the induction step, assume that $n \geq 0$ and the claim holds for $n$. We want to prove it for $n+1$. We use the mean value theorem as you suggest: $$\frac{a_{n+1} - x_0}{a_n - x_0} = \frac{f(a_n) - f(x_0)}{a_n - x_0} = f'(c_n)$$ where $c_n$ lies between $a_n$ and $x_0$. Since $a_n$ and $x_0$ both lie in $I$, this implies that $c_n\in I$ as well, and thus by assumption that $|f'(c_n)| \leq r$. We conclude that $$\left|\frac{a_{n+1} -x_0}{a_n - x_0}\right|\leq r$$ and thus by induction $$|a_{n+1}-x_0| \leq r|a_n - x_0| \leq r^{n+1}|a - x_0|.$$ This establishes (2.). To prove (1.), we use what we just proved: the distance from $a_{n+1}$ to $x_0$ is less than the distance of $a_n$ to $x_0$. Since $a_n\in I$ and $I$ is symmetric around $x_0$, this implies $a_{n+1}\in I$ as well. Now we're done. 
