# Prove that limit doesn't exist

• Feb 8th 2010, 05:40 PM
Pinkk
Prove that limit doesn't exist
Sorry for all the questions but if there is a recurrence relation such that $\displaystyle x_{1} =1$, and for $\displaystyle n \ge 1, x_{n+1} = 3x_{n}^{2}$, how do I show that $\displaystyle \lim_{n\to \infty}\,(x_{n})$ does not exist. It is obvious, but I do not know how to show it. I'm assuming there is a general formula for this recurrence relation but I cannot find it. Or is there another way to prove it altogether?
• Feb 8th 2010, 06:38 PM
Drexel28
Quote:

Originally Posted by Pinkk
Sorry for all the questions but if there is a recurrence relation such that $\displaystyle x_{1} =1$, and for $\displaystyle n \ge 1, x_{n+1} = 3x_{n}^{2}$, how do I show that $\displaystyle \lim_{n\to \infty}\,(x_{n})$ does not exist. It is obvious, but I do not know how to show it. I'm assuming there is a general formula for this recurrence relation but I cannot find it. Or is there another way to prove it altogether?

You could just solve the recurrence relation: $\displaystyle x_{n+1}=3x_n^2=3\left(3x_{n-1}^2\right)^2=\cdots=3^{2^{n-1}-1}x_1=3^{2^{n-1}-1}$. Or, let $\displaystyle \ell_n=x_n-n$. Clearly we have that $\displaystyle \ell_1=1-1\geqslant 0$. Now, suppose that $\displaystyle \ell_n\geqslant 0$. Then, $\displaystyle \ell_{n+1}=x_{n+1}-(n+1)=3x_n^2-(n+1)$$\displaystyle \geqslant 2x_n^2-2n\geqslant 2\left(x_n^2-n^2\right)=2\left(x_n-n\right)\cdot\left(x_n+n\right)\geqslant 3\cdot0\cdot 1=0$. It follows by induction that $\displaystyle \ell_n\geqslant 0\implies x_n\geqslant n$.
• Feb 8th 2010, 06:50 PM
Pinkk
Thanks. It makes sense now but I am no good at solving recurrence relations. But I do not believe I needed to do so anyway. I just figured out that by the recurrence relation, if the limit did exist, it'd have to be either 0 or 1/3. But since every term of that sequence is greater than or equal to 1, then the limit must also be greater than or equal to 1, so obviously it does not exist.