Answer
(a) this function is odd.
(b) $x=\pm 2\pi, \pm 4\pi, \pm 6\pi ... \pm 2n\pi, n=1,2,3 ... $
(c) The graph of the function is shown in the figure.
(d) As $x\to -\infty, f(x)\to 0$ and as $x\to \infty, f(x)\to 0$
(e) $x\to0, f(x)\to 0$
Work Step by Step
(a) $f(-x)=\frac{1-cos(-x)}{-x}=-\frac{1-cos(x)}{x}=-f(x)$, so this function is odd.
(b) Let $cos(x)=1, x\ne0$, we can find the x-intercepts as
$x=\pm 2\pi, \pm 4\pi, \pm 6\pi ... \pm 2n\pi, n=1,2,3 ... $
(c) The graph of the function is shown in the figure.
(d) As $x\to -\infty, f(x)\to 0$ and as $x\to \infty, f(x)\to 0$
(e) As $x\to 0$, both the numerator and the denominator approach zero,
and the ratio approaches zero ($x\to0, f(x)\to 0$) as shown in the graph.
