Solution to Elementary Analysis: The Theory of Calculus Second Edition Section 9 Exercise 9.6

Solution:

### Part a

This part is the same as that of Exercise 9.4 and Exercise 9.5. One needs to solve

$$

a=3a^2

$$ for $a$. Hence $a=0$ or $a=\dfrac{1}{3}$.

### Part b

The sequence $x_1$ diverges to $+\infty$. Indeed, one can show that $x_n\ge 3^{n-1}$ by induction and hence $\lim x_n=+\infty$.

We show $x_n\ge 3^{n-1}$ by induction. The base case is clear.

Suppose $x_n\ge 3^{n-1}$, we show that $x_{n+1}\ge 3^{n}$. This follows from

$$

x_{n+1}=3x_n^2\ge 3(3^{n-1})^2=3\cdot 3^{2n-2}=3^{2n-1}\ge 3^n.

$$ Hence we have $x_n\ge 3^{n-1}$ for all $n$.

Now it is clear that $\lim x_n$ is infinity.

### Part c

There is no contradiction at all. Part (a) is obtained from the assumption that $\lim x_n$ exists. However, we can show that $\lim x_n$ does not exist in Part (b). Hence in reality, we cannot obtain $\lim x_n=\dfrac{1}{3}$ or $\lim x_n=0$.

This exercise explains if we would like to find limit using the method used in Exercise 9.4 and Exercise 9.5. We must show convergence first. Otherwise, such a method cannot be applied at all.