Tangent plane approximation: $f_1(x_0,y_0)h+f_2(x_0,y_0)k + \epsilon(h,k)\sqrt{h^2 + k^2}$ can be rewritten with $|h|+|k|$ instead of $\sqrt{h^2+k^2}$

Question

As a condition for a function $f:\mathbb{R}^2 \to \mathbb{R}$ to be differentiable at $(x_0, y_0)$, the author asks that it be possible to write $$f(x_0 +h, y_0+k) - f(x_0, y_0)=f_1(x_0,y_0)h+f_2(x_0,y_0)k + \epsilon(h,k)\sqrt{h^2 + k^2}$$ where $\epsilon(h,k) \to 0$ as $\sqrt{h^2 + k^2} \to 0$.

Notice that this is comes from the tangent plane function $$T(x,y) = f(x_0,y_0) + f_1(x_0,y_0)(x-x_0) + f_2(x_0,y_0)(y-y_0)$$ which approximates $f(x)$ at $(x_0, y_0)$; and also from the corresponding relative error $\frac{|f(x,y) - T(x,y)|}{\sqrt{(x-x_0)^2 + (y-y_0)^2}}$ which goes to $0$ as $(x,y)$ approaches $(x_0, y_0)$.

The author offers the alternative condition $$f(x_0 + h, y_0 +k) - f(x_0, y_0) = f_1(x_0, y_0)h + f_2(x_0,y_0)k + \epsilon(h,k)(|h| + |k|)$$ by observing that $\sqrt{h^2 + k^2} \leq |h| + |k| \leq \sqrt{2}\sqrt{h^2 +k^2}$.

I'm not sure, but I think that the substitution is justified because the initial error function: $$\epsilon(h,k)= \frac{f(x_0 +h, y_0+k) - f(x_0, y_0) - f_1(x_0,y_0)h-f_2(x_0,y_0)k}{\sqrt{h^2 + k^2}} $$ is bigger than the resulting error function: $$\epsilon(h,k)= \frac{f(x_0 +h, y_0+k) - f(x_0, y_0) - f_1(x_0,y_0)h-f_2(x_0,y_0)k}{|h| + |k|} $$

My question is: why does $|h| + |k|$ have to have the upper bound $\sqrt{2}\sqrt{h^2 + k^2}$?

This question comes from reading page 711 of Elementary Real Analysis by Thomson, Bruckner and Bruckner.

Answer 1

$(|h|+|k|)^2 = h^2+k^2+2|hk|$

For any 2 real numbers $a,b,~~(a-b)^2 \ge0 \Rightarrow \boxed{a^2+b^2\ge2ab}$.

So,

$(|h|+|k|)^2 = h^2+k^2+2|hk| \le h^2+k^2+ h^2+k^2 = 2(h^2+k^2)\Rightarrow \boxed{|h|+|k|\le\sqrt2\sqrt{h^2+k^2}}$

Answer 2

The inequality is derived using $2ab \leq a^2 + b^2$ for $a,b \in \Bbb R$ $$(|h| + |k|)^2 = |h|^2 + |k|^2 + 2|h||k| \leq 2(|h|^2 + |k|^2)$$ Taking the square root we get $$|h| + |k| \leq \sqrt{2}\sqrt{|h|^2 + |k|^2}$$