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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 $\lim x_n=0$.