I need to find the range of the function, $$f(x)= \log_{[x-1]}\sin x$$ where $[.]$ denotes the greatest integer function of $x$.
Now, here's what I tried, but I got stuck big time.
Firstly, I know that $-1 \leq \sin x \leq 1$
But, for the logarithm here to be defined on the reals, I have $\sin x>0$, so finally I get that $1 \geq \sin x >0$
Also, for the base of the logarithm to be valid, I get $[x-1]>0$ and $[x-1] \ne 1$
So, $[x] >1 \implies x \geq 2$ and $[x] \ne 2$. SO, finally, $x > 2$. So, $[x]$ can take values in the interval $[3,\infty)$ and thus $[x-1]$ can take values in the interval $[2,\infty)$
Now, I get the fact that as the base of the logarithm is greater than $1$, so it is clearly an increasing function, and also, for $\sin x = 1$, I get the value of $f(x)= 0$
But how do I find the lower bound for the range? The book states the answer to be $(-\infty,0]$, but that $-\infty$ can only occur if $\sin x \to0$ right? Can anyone help me with a detailed explanation, I have been stuck on this problem for hours. Help would be much appreciated.
For this to be defined, $[x-1]>1 \implies x\ge3$. For $x \in [3,4)$, the function becomes $\log_2(\sin x)$. Here, $\sin x \to 0$ as $x \to \pi \approx 3.141592654....$. Thus, the lower bound would be $-\infty$. You have already got the maximum value as zero in your answer, thus the range is $(-\infty, 0]$