Show that the limit is divergent

Question

I want to calculate the following limit: $$ \lim_{x \rightarrow \infty} Ax^{b} \cos(c \log(x)), $$ where $A$, $b>0$ and $c$ are some constants.

I suppose that the function $Ax^{b} \cos(c \log(x))$ is not convergent when $x\rightarrow \infty$.

Firstly, I think I can show that $$ A\cos(c \log(x)) $$ is not convergent when $x\rightarrow \infty$.

For this purpose I use two different subsequences and show that the limit converges to different values:

1. $x = e^{\frac{2n}{c}\pi}$, then $\lim_{x \rightarrow \infty} A \cos(c \log(x)) = A$.

2. $x = e^{\frac{2n+1}{c}\pi}$, then $\lim_{x \rightarrow \infty} A \cos(c \log(x)) = -A$.

Of course, I assumed that $A\neq 0$.

Moreover, I know that: $$ \lim_{x \rightarrow \infty} x^{b} = +\infty $$ for $b>0$

Then, I claim that: $$ Ax^{b} \cos(c \log(x)) $$ is not convergent when $x\rightarrow \infty$, because it is a product of a function which is not convergent and a function which has a limit equal to $\infty$.

Is this solution correct?

Answer 1

Yes it is correct, we can simply note that since $|c \log x|\to \infty$ is continuos

• $\cos(c \log(x))$ oscillates from $-1$ and $1$

and

• $|Ax^{b}|\to \infty$

then

• $Ax^{b} \cos(c \log(x))$ doesn't converges ("oscillates between" $+\infty$ and $-\infty$)

Answer 2

The fact that $Ax^b$ diverges is unimportant. It suffices to say that the function is alternating and doesn't tend to $0$.