Showing a function of three variables is differentiable at a point

Question

I am having trouble showing that the following function $f: \mathbb{R}^3 \rightarrow \mathbb{R}$ given by $$f(x,y,z) = xy + yz + xz$$ is differentiable at $(1,1,1)$. I have used the definition of differentiability along with the Jacobian to reduce the problem to showing that the limit $$\lim_{\textbf{x}\to (1,1,1)} \frac{|xz + xy + yz + 3 - 2(x+y+z)|}{\sqrt{(x-1)^2 + (y-1)^2 + (z-1)^2}} = 0$$ I have tried using the pinching principle, but could not find a useful upper bound. I also tried to use the $\epsilon$-$\delta$ definition of limits, but could not get anything useful. How does one show that this limit is equal to $0$?

Answer 1

I will suggest to use the version that \begin{align*} \lim_{(h,k,l)\rightarrow(0,0,0)}\dfrac{f(h,k,l)-f(1,1,1)-(2h+2k+2l)}{\sqrt{h^{2}+k^{2}+l^{2}}}. \end{align*} After simplifying, we get the inner expression as \begin{align*} \dfrac{hk+kl+lh}{\sqrt{h^{2}+k^{2}+l^{2}}}, \end{align*} we now use the Cauchy-Schwarz inequality that \begin{align*} |hk+kl+lh|\leq\sqrt{h^{2}+k^{2}+l^{2}}\sqrt{k^{2}+l^{2}+h^{2}} \end{align*} to get \begin{align*} \left|\dfrac{hk+kl+lh}{\sqrt{h^{2}+k^{2}+l^{2}}}\right|\leq\sqrt{h^{2}+k^{2}+l^{2}}<\epsilon \end{align*} by choosing $\delta=\epsilon$.