Try this problem 20 from TIFR 2013 based on Sequence Limit – Fixed Point. This problem is an application of the Banach Fixed Point Theorem.

Question: TIFR 2013 problem 20


Consider the function \(f(x)=ax+b\) with \(a,b\in\mathbb{R}\). Then the iteration \(x_{n+1}=f(x_n)\); \(n\ge0\) for a given \(x_0\) converges to \(\frac{b}{1-a}\) whenever \(0<a<1\).


Banach Fixed Point Theorem


Once the existence of limit is guaranteed, we can safely calculate the limit. If limit is \(x\) then \( x=ax+b\). In other words, \(x=\frac{b}{1-a}\).

The question really is does limit exists?

\(x_{n+1}-x_{n}=f(x_{n})-f(x_{n-1}) \)


For \(n<m\),

\(x_m-x_n=x_m-x_{m-1}+x_{m-1}-x_{m-2}+…-x_{n} \)

\( =(a^{m}+…a^{n})(x_1-x_0)=\frac{a^n}{1-a}(x_1-x_0) \)

And the right hand side converges to 0 as n tends to infinity (Here,we are using the fact that \(0<a<1\)) i.e, the right hand side can be made arbitrarily small using large enough n, so \(x_n\) is a Cauchy sequence. We are in \(\mathbb{R}\), so by completeness, \(x_n\) converges.


The above discussion really didn’t make use of Banach Fixed Point Theorem. But knowledge is power. And if known that :

“Let \(T:X\to X\) be a contraction mapping, X-complete metric space. Then T has precisely one fixed point \(u\in X\). Furthermore, for any \(x\in X\) the sequence \(T^k (x)\) converges and the limit is \(u\). ” (Banach Fixed Point Theorem).

the answer is immediate. The above solution runs in the line of the proof of this Theorem.

Some Useful Links: