I have some limits I am supposed to find and I'm having trouble understanding the question.
Given:
$$f(x) = \begin{cases} 2 & \text{ if } \left \lceil x \right \rceil \text{is even}\\-1 & \text{ if } \left \lceil x \right \rceil \text{is odd} \end{cases}$$
$g(x) = f\left(\frac{1}{x}\right)$
what does $g(x) = f\left(\frac{1}{x}\right)$ mean? (maybe I'm making it more complicated than it should be in my head, it's supposed to be a 'challenge' problem.)
example: $g\left(\frac{1}{3}\right)$, does that mean $g\left(\frac{1}{\frac{1}{3}}\right)$ so then $g(3) = -1$ ?
or like a composite function where $g\left(\frac{1}{3}\right)$ is odd so $f\left(\frac{1}{3}\right) = -1$ then $g(x) = \frac{1}{-1} = -1$?
I realize both come out to $-1$, but I think it makes a difference for when I have to find the limits. Or maybe not?
$g(x)=f(\frac 1x)$ is just as you suspect-if I ask you for $g(x)$ you compute $y=\frac 1x$ and ask somebody for $f(y)$. For any $x \gt 1, 0 \lt \frac 1x \lt 1, \lceil \frac 1x \rceil=1$, so $f(\frac 1x)=-1$.