How do I evaluate: $\limsup_{x\to0-}\frac{\sqrt{-x\sin^2(\ln |x|)}}{x}$?

Let $f:\mathbb R\to\mathbb R$ be defined by $$f(x)=\begin{cases}x[1+\sin(\ln x)]&\text{if }x>0\\0&\text{if }x=0\\x+\sqrt{-x\sin^2(\ln |x|)}&\text{if }x<0.\end{cases}$$ 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{align} D^-f(0)&=\limsup\limits_{h\to0-}\frac{f(h)-f(0)}h\\\\ &=1+\limsup\limits_{h\to0-}\frac{\sqrt{-h\sin^2(\ln |h|)}}h. \end{align} Next putting: $-h=p$ so that $p\to0+,$ I obtained: $$D^-f(0)=1-\liminf\limits_{p\to0+}\frac{|\sin(\ln |p|)|}{\sqrt p}.$$ 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.

Solution 1:

We have $$\begin{align}D^{-}f(0)&=\limsup_{h\to 0^{-}}\frac{h+\sqrt{-h\sin^2(\ln(-h))}}{h}\\ &=\limsup_{h\to 0^{-}}\left(1-\sqrt{-\frac{\sin^2(\ln(-h))}{h}}\right)\end{align} $$ In the last line, we used $\frac 1h=-\sqrt{\frac{1}{h^2}}$ for $h<0$. Let $$g(h)=1-\sqrt{-\frac{\sin^2(\ln(-h))}{h}} $$ Clearly, $g(h)\le 1$ for all $h<0$. The limit superior would then be equal to $1$ if, for example, we show $g$ attains the value of $1$ in every left neighborhood of $0$. Consider the sequence $$x_n=-e^{-n\pi} $$ Note $x_n<0$, $\lim_{n\to\infty}x_n=0$, and $$g(x_n)=1-\sqrt{e^{n\pi}\sin^2(-n\pi)}=1-0=1 $$ Thus, $D^{-}f(0)=1$.