$\lim\limits_{h\to(0,0)}\frac{\sqrt[3]{h_1h_2+h_2+h_1+1}-1-\frac{h_1}{3}-\frac{h_2}{3}}{\sqrt{h_1^2+h_2^2}}$ 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.
 A: 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 $(h_1,h_2)\to 0.$
