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I am trying to show that a function has a total derivative at $(1,1)$ but I got stuck trying to show $$\lim\limits_{h\to(0,0)}\frac{\sqrt[3]{h_1h_2+h_2+h_1+1}-1-\frac{h_1}{3}-\frac{h_2}{3}}{\sqrt{h_1^2+h_2^2}}=0$$ Where $h=(h_1,h_2)\in\mathbb{R}^2$. Since $h$ approaches $(0,0)$ $h_1$ and $h_2$ have to approach $0$ aswell but I am relatively new to multivariable limits so I need some help.

Zacharias Zarowski
  • 335
  • Well, you clearly have to reduce the root, so try the formula $x-y = \frac{x^3-y^3}{x^2+xy+y^2}$ – Jakobian Jul 04 '19 at 20:19
  • @Jakobian Sorry but I don't see how I can utilize that formula here :( – Zacharias Zarowski Jul 04 '19 at 20:29
  • You use the formula (with $x = \sqrt[3]{h_1h_2+h_1+h_2+1}, \ y = 1+h_1/3+h_2/3$), the denominator is something converging to $3$, the nominator you expand. Then you reduce the terms in nominator by identifying the ones which are multiples of $h_1^2+h_2^2$. I've got something like $h_1h_2$ multiplied by a polynomial in the end. Well, $\frac{h_1h_2}{\sqrt{h_1^2+h_2^2}}\to 0$ you can show by some easy inequalities. – Jakobian Jul 04 '19 at 20:43
  • @Jakobian I can't type the fraction here to show you what I've got so far but I am stuck at reducing the nominator... I know how to show $\frac{h_1h_2}{\sqrt{h_1^2+h_2^2}}\to 0$ but could you help me again? (I know this is probably very easy for you but it's very confusing for me) – Zacharias Zarowski Jul 04 '19 at 21:08
  • You should get $$-\frac{1}{27}[h_1^3+h_2^3+3(h_1^2h_2+h_1h_2^2)+9(h_1^2+h_2^2)-9h_1h_2].$$ Note that $h_2^3+h_1^2h_2 = h_2(h_1^2+h_2^2)$ though, so those we can ignore and we're left with $-\frac{1}{27}(2(h_1^2h_2+h_1h_2^2)-9h_1h_2) = -\frac{1}{27}h_1h_2[2h_1+2h_2-9]$ – Jakobian Jul 04 '19 at 21:22
  • @Jakobian Got it! But I don't understand the last step. I get $$-\frac{1}{27}(h_1^3+2h_1^2h_2+9h_1^2+3h_1h_2^2-9h_1h_2+9h_2^2+h_2(h_1^2+h_2^2))$$ – Zacharias Zarowski Jul 04 '19 at 22:02
  • Yeah, but $h_1^3+h_1h_2^2 = h_1(h_1^2+h_2^2)$ as well. – Jakobian Jul 04 '19 at 22:07
  • So $-\frac{1}{27}(9(h_1^2+h_2^2)+2h_1^2h_2+2h_1h_2^2-9h_1h_2+(h_1+h_2)(h_1^2+h_2^2))$? – Zacharias Zarowski Jul 04 '19 at 22:13
  • Yes, but $2h_1^2h_2+2h_1h_2^2-9h_1h_2 = h_1h_2(2h_1+2h_2-9)$ also. – Jakobian Jul 04 '19 at 22:17
  • @Jakobian Now all in all I get $$\frac{-\frac{\sqrt{h_1^2+h_2^2}}{27}((9+h_1+h_2)(h_1^2+h_2^2)+h_1h_2(2h_1+2h_2-9))}{(h_1h_2+h_1+h_2+1)^\frac{2}{3}+\sqrt[3]{h_1h_2+h_1+h_2+1}(1+\frac{h_1}{3}+\frac{h_2}{3})+(1+\frac{h_1}{3}+\frac{h_2}{3})^2}$$ How can I simplify that? Also I shouldnt leave as many comments but I don't have enough reputation to chat apparently – Zacharias Zarowski Jul 04 '19 at 22:30
  • In the nominator there, it should be $$-\frac{\sqrt{h_1^2+h_2^2}}{27}(9+h_1+h_2)-\frac{h_1h_2}{27\sqrt{h_1^2+h_2^2}}(2h_1+2h_2-9)$$ and we're pretty much set, the denominator goes to $3$, and nominator to $0$. (first term in nominator converges to $0$, and the second one is something converging to $0$ times bounded, so converges to $0$). – Jakobian Jul 04 '19 at 22:38
  • But $$-\frac{\sqrt{h_1^2+h_2^2}}{27}(9+h_1+h_2)(h_1^2+h_2^2)-\frac{\sqrt{h_1^2+h_2^2}}{27}h_1h_2(2h_1+2h_2-9)\neq-\frac{\sqrt{h_1^2+h_2^2}}{27}(9+h_1+h_2)-\frac{h_1h_2}{27\sqrt{h_1^2+h_2^2}}(2h_1+2h_2-9)$$ or am I not seeing something? Also should I write an answer to this question now? – Zacharias Zarowski Jul 04 '19 at 22:56
  • It's not, left side differs from the right by a factor of $h_1^2+h_2^2$. I'm not completely sure about politics of this site, so I won't help with the second question. Perhaps you should. – Jakobian Jul 04 '19 at 23:05
  • Ok I will. hopefully it's ok. Thanks for all!!! – Zacharias Zarowski Jul 04 '19 at 23:07

2 Answers2

1

We have \begin{align*} \lim\limits_{h\to(0,0)}&\frac{\sqrt[3]{h_1h_2+h_2+h_1+1}-1-\frac{h_1}{3}-\frac{h_2}{3}}{\sqrt{h_1^2+h_2^2}}\\ =\lim\limits_{h\to(0,0)}&\frac{(h_1h_2+h_1+h_2+1-\overbrace{(1+\frac{h_1}{3}+\frac{h_2}{3})^3}^{=\frac{(3+h_1+h_2)^3}{27}})}{((h_1h_2+h_1+h_2+1)^\frac{2}{3}+\sqrt[3]{h_1h_2+h_1+h_2+1}(1+\frac{h_1}{3}+\frac{h_2}{3})+(1+\frac{h_1}{3}+\frac{h_2}{3})^2)\sqrt{h_1^2+h_2^2}}\\ =\lim\limits_{h\to(0,0)}&\frac{(h_1h_2+h_1+h_2+1-\frac{1}{27}(h_1^3+3h_1^2h_2+9h_1^2+3h_1h_2^2+18h_1h_2+27h_1+h_2^3+9h_2^2+27h_2+27))}{((h_1h_2+h_1+h_2+1)^\frac{2}{3}+\sqrt[3]{h_1h_2+h_1+h_2+1}(1+\frac{h_1}{3}+\frac{h_2}{3})+(1+\frac{h_1}{3}+\frac{h_2}{3})^2)\sqrt{h_1^2+h_2^2}}\\ =\lim\limits_{h\to(0,0)}&\frac{-\frac{1}{27}(h_1^3+3h_1^2h_2+9h_1^2+3h_1h_2^2-9h_1h_2+h_2^3+9h_2^2)}{((h_1h_2+h_1+h_2+1)^\frac{2}{3}+\sqrt[3]{h_1h_2+h_1+h_2+1}(1+\frac{h_1}{3}+\frac{h_2}{3})+(1+\frac{h_1}{3}+\frac{h_2}{3})^2)\sqrt{h_1^2+h_2^2}}\\ =\lim\limits_{h\to(0,0)}&\frac{-\frac{1}{27}(9(h_1^2+h_2^2)+h_1h_2(2h_1+2h_2-9)+(h_1+h_2)(h_1^2+h_2^2))}{((h_1h_2+h_1+h_2+1)^\frac{2}{3}+\sqrt[3]{h_1h_2+h_1+h_2+1}(1+\frac{h_1}{3}+\frac{h_2}{3})+(1+\frac{h_1}{3}+\frac{h_2}{3})^2)\sqrt{h_1^2+h_2^2}}\\ =\lim\limits_{h\to(0,0)}&\frac{\overbrace{-\frac{\sqrt{h_1^2+h_2^2}}{27}(9+h_1+h_2)-\frac{h_1h_2}{27\sqrt{h_1^2+h_2^2}}(2h_1+2h_2-9)}^{\to0}}{\underbrace{(h_1h_2+h_1+h_2+1)^\frac{2}{3}+\sqrt[3]{h_1h_2+h_1+h_2+1}(1+\frac{h_1}{3}+\frac{h_2}{3})+(1+\frac{h_1}{3}+\frac{h_2}{3})^2}_{\to3}}=0 \end{align*}

$\frac{h_1h_2}{\sqrt{h_1^2+h_2^2}}\to 0$ has already been shown here: https://math.stackexchange.com/q/417052

Zacharias Zarowski
  • 335
1

Let $f(z)=(1+z)^{1/3}.$ When $z\ne -1$ we have $f(z)=f(0)+zf'(0)+z^2 f''(y)/2$ for some $y$ such that $|y|\le |z|.$ We have $f(0)=1 $ and $f'(0)=1/3.$

Let $z=h_1+h_2+h_1h_2.$ When $(h_1,h_2)\ne (0,0)$ and $z \ne -1$ we have $$\frac {(1+h_1+h_2+h_1h_2)^{1/3}-1-h_1/3-h_2/3}{\sqrt {h_1^2+h_2^2}}=$$ $$=\frac {h_1h_2/3+z^2f''(y)/2} {\sqrt {h_1^2+h_2^2}}.$$ Let $h=\min (|h_1|,|h_2|)$ and $H=\max (|h_1|,|h_2|)$. We have $\sqrt {h_1^2+h_2^2}\ge H.$

(i). We have $|h_1h_2|=hH.$ Therefore $$\left|\frac {h_1h_2/3}{\sqrt {h_1^2+h_2^2}}\right|\le \frac {hH/3}{H}=h/3.$$ And $h\to 0$ as $(h_1,h_2)\to 0.$

(ii). When $H<1$ we have $|z|\le |h_1|+|h_2|+|h_1|\cdot |h_2|\le H+H+H^2<3H$. Therefore when $0<H<1$ we have $$\left|\frac {z^2f''(y)/2}{\sqrt {h_1^2+h_2^2}}\right|\le \frac {(3H)^2\cdot|f''(y)/2|}{H}=(9H/2)\cdot|f''(y)|.$$ Now note that as $(h_1,h_2)\to 0$ we have $ H\to 0$ and $z\to 0.$ And $|y|\le |z|$ so $ y\to 0$ also. Finally $f''(y)$ is continuous at $y=0$ so $|f''(y)|\to |f''(0)|=2/9$ as $(h_1,h_2)\to 0.$

DanielWainfleet
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