Let f : R → R be defined by f ( x ) = {...

Let $f:\mathbb{R}\to \mathbb{R}$ be defined by
$f(x)=\{\begin{array}{ll}x[1+\sin (\ln x)]& \text{if }x>0\\ 0& \text{if }x=0\\ x+\sqrt{-x{\sin }^2(\ln |x|)}& \text{if }x<0.\end{array}$
Find the values of the four Dini derivatives of $f$ at $x=0.$
I could easily find out the values of the upper right $D^{+}f$ and lower right $D_{+}f$ Dini derivatives at $x=0,$ values being 2 and 0 respectively. But, in case of upper left Dini derivative of $f$ at $x=0,$ we have:
$\begin{array}{rl}{D}^{-}f(0)& =\underset{h\to 0-}{lim sup}\frac{f(h)-f(0)}{h}\\ \\ & =1+\underset{h\to 0-}{lim sup}\frac{\sqrt{-h{\sin }^2(\ln |h|)}}{h}.\end{array}$
Next putting: $-h=p$ so that $p\to 0+,$, I obtained:
$D^{-}f(0)=1-\underset{p\to 0+}{lim inf}\frac{|\sin (\ln |p|)|}{\sqrt{p}}.$
Since it is limit inferior and there is a modulus of sine function present inside, so can we take the sine part as 0 ? This way the answer would match but I feel like I'm avoiding the effect of the term $\sqrt{p}$ present in the denominator. I am confused. Please give some insights. Thanks in advance.