Exercise 5.2.7

We need $a>0$ to make $g_a$ continuous at zero, for differentiation notice

Thus for differentiation we need $x^{a-1}\sin <!--\; FUNCTION\; APPLICATION\; -->(1∕x)$ to have a limit at zero. Keep this in mind for the problems. Also note

(a)

To get unboundedness we need the $x^{a-2}\cos <!--\; FUNCTION\; APPLICATION\; -->(1∕x)$ term to become unbounded, so pick $a=1.5$ to satisfy $a>1$ (differentiable) and $a<2$ (unbounded derivative)

(b)

Pick $a=2.5$, It’s clear that $g_a^{\prime }(x)=2.5x^{1.5}\sin <!--\; FUNCTION\; APPLICATION\; -->(1∕x)-x^{0.5}\cos <!--\; FUNCTION\; APPLICATION\; -->(1∕x)$ is continuous at zero, but not differentiable since $x^{0.5}\cos <!--\; FUNCTION\; APPLICATION\; -->(1∕x)$ isn’t differentiable at zero and $2.5x^{1.5}\sin <!--\; FUNCTION\; APPLICATION\; -->(1∕x)$ is (if both terms were not differentiable they could cancel eachother out, constant vigilance!).

(c)

Pick $a=3.5$, for all intents and purposes $g_a^{\prime }$ behaves like $g_{a-2}$ (because the $x^{a-2}\cos <!--\; FUNCTION\; APPLICATION\; -->(1∕x)$ term acts as a bottleneck) meaning $g_a^{″}$ exists, but isn’t continuous since we get another $-2$ to $a$ leading to a $x^{-0.5}$ term in $g_a^{″}$.