Villaretq0

2022-06-22

Let $f:\mathbb{R}\to \mathbb{R}$ be defined by

$f(x)=\{\begin{array}{ll}x[1+\mathrm{sin}(\mathrm{ln}x)]& \text{if}x0\\ 0& \text{if}x=0\\ x+\sqrt{-x{\mathrm{sin}}^{2}(\mathrm{ln}|x|)}& \text{if}x0.\end{array}$

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:

$\begin{array}{rl}{D}^{-}f(0)& =\underset{h\to 0-}{lim\u2006sup}\frac{f(h)-f(0)}{h}\\ \\ & =1+\underset{h\to 0-}{lim\u2006sup}\frac{\sqrt{-h{\mathrm{sin}}^{2}(\mathrm{ln}|h|)}}{h}.\end{array}$

Next putting: $-h=p$ so that $p\to 0+,$, I obtained:

${D}^{-}f(0)=1-\underset{p\to 0+}{lim\u2006inf}\frac{|\mathrm{sin}(\mathrm{ln}|p|)|}{\sqrt{p}}.$

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.

$f(x)=\{\begin{array}{ll}x[1+\mathrm{sin}(\mathrm{ln}x)]& \text{if}x0\\ 0& \text{if}x=0\\ x+\sqrt{-x{\mathrm{sin}}^{2}(\mathrm{ln}|x|)}& \text{if}x0.\end{array}$

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:

$\begin{array}{rl}{D}^{-}f(0)& =\underset{h\to 0-}{lim\u2006sup}\frac{f(h)-f(0)}{h}\\ \\ & =1+\underset{h\to 0-}{lim\u2006sup}\frac{\sqrt{-h{\mathrm{sin}}^{2}(\mathrm{ln}|h|)}}{h}.\end{array}$

Next putting: $-h=p$ so that $p\to 0+,$, I obtained:

${D}^{-}f(0)=1-\underset{p\to 0+}{lim\u2006inf}\frac{|\mathrm{sin}(\mathrm{ln}|p|)|}{\sqrt{p}}.$

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.

Savanah Hernandez

Beginner2022-06-23Added 16 answers

We have

$\begin{array}{rl}{D}^{-}f(0)& =\underset{h\to {0}^{-}}{lim\u2006sup}\frac{h+\sqrt{-h{\mathrm{sin}}^{2}(\mathrm{ln}(-h))}}{h}\\ & =\underset{h\to {0}^{-}}{lim\u2006sup}(1-\sqrt{-\frac{{\mathrm{sin}}^{2}(\mathrm{ln}(-h))}{h}})\end{array}$

In the last line, we used $\frac{1}{h}=-\sqrt{\frac{1}{{h}^{2}}}$ for $h<0$. Let

$g(h)=1-\sqrt{-\frac{{\mathrm{sin}}^{2}(\mathrm{ln}(-h))}{h}}$

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 ${x}_{n}<0$, $\underset{n\to \mathrm{\infty}}{lim}{x}_{n}=0$, and

$g({x}_{n})=1-\sqrt{{e}^{n\pi}{\mathrm{sin}}^{2}(-n\pi )}=1-0=1$

Thus, ${D}^{-}f(0)=1$.

$\begin{array}{rl}{D}^{-}f(0)& =\underset{h\to {0}^{-}}{lim\u2006sup}\frac{h+\sqrt{-h{\mathrm{sin}}^{2}(\mathrm{ln}(-h))}}{h}\\ & =\underset{h\to {0}^{-}}{lim\u2006sup}(1-\sqrt{-\frac{{\mathrm{sin}}^{2}(\mathrm{ln}(-h))}{h}})\end{array}$

In the last line, we used $\frac{1}{h}=-\sqrt{\frac{1}{{h}^{2}}}$ for $h<0$. Let

$g(h)=1-\sqrt{-\frac{{\mathrm{sin}}^{2}(\mathrm{ln}(-h))}{h}}$

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 ${x}_{n}<0$, $\underset{n\to \mathrm{\infty}}{lim}{x}_{n}=0$, and

$g({x}_{n})=1-\sqrt{{e}^{n\pi}{\mathrm{sin}}^{2}(-n\pi )}=1-0=1$

Thus, ${D}^{-}f(0)=1$.