# Please explain me how can I show that the last limit does not exist? [closed]

I posted my answer with its question.

But how can I show that the last limit -on the second page- does not exist?

That is,

## $\mathbf{\lim_{(h_1, h_2)\to (0,0)}\frac{\sqrt {|h_1.h_2|}}{\sqrt {h_1^2,+ h_2^2}}}$ does not exist. How?

11.2.3. Prove that $f(x,y)=\sqrt{|xy|}$ is not differentiable at $(0,0)$.

Solution.

By a theorem, we need to check that $f$ has first order partial derivatives and they are continuous at $(0,0)$.

$$\frac{\partial f}{\partial x}(0,0) = f_x(0,0) =\lim\limits_{h\to0} \frac{f((0,0)+h(1,0))-f(0,0)}h = \lim\limits_{h\to0} \frac{f(h,0)-f(0,0)}h = \lim\limits_{h\to0} \frac{\sqrt{|h\cdot0|}}h=0$$

$$\frac{\partial f}{\partial y}(0,0) = f_y(0,0) =\lim\limits_{h\to0} \frac{f((0,0)+h(0,1))-f(0,0)}h = \lim\limits_{h\to0} \frac{f(h,0)-f(0,0)}h = \lim\limits_{h\to0} \frac{\sqrt{|0\cdot h|}}h=0$$

Consequently, $\delta f(0,0)=(0,0)$, i.e., $f$ has first-order partial derivatives at zero.

Let's show the continuity of first order partial derivatives of $f$ at $(0,0)$.

$$\lim\limits_{(h_1,h_2)\to(0,0)} \frac{f((0,0)+(0,0)\cdot(h_1,h_2))-[f(0,0)+\partial f(0,0)(h_1,h_2))]}{\|(h_1,h_2)\|}= \lim\limits_{(h_1,h_2)\to(0,0)} \frac{f(h_1,h_2)-f(0,0)-Df(0,0)(h_1,h_2)}{\|(h_1,h_2)\|} = \lim\limits_{(h_1,h_2)\to(0,0)} \frac{f(h_1,h_2)}{\|(h_1,h_2)\|} = \lim\limits_{(h_1,h_2)\to(0,0)} \frac{\sqrt{|h_1\cdot h_2|}}{\sqrt{h_1^2+h_2^2}} =$$

## closed as off-topic by Xander Henderson, GNUSupporter 8964民主女神 地下教會, Did, HK Lee, VladApr 17 '18 at 12:04

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• I did my best to retype the text from your picture. There were some places where I had problems with reading your text - when you have time, please, check whether this is what you intended to write. – Martin Sleziak Apr 14 '18 at 16:20

Try picking $h_1=h_2$, say. That is, approach the origin through the line $x=y$. What is the limit? Is it $0$? Now look at $h_1=2h_2$.
• Not $0$. the limit is equal to $\frac{1}{\sqrt 2}$ when $h_1 =h_2$ – user315 Jul 9 '13 at 0:05
• Then, $\frac{\sqrt 2}{\sqrt 5}$ when $h_1=2 h_2$ – user315 Jul 9 '13 at 0:07
• (+1) you strongly ruled the line $y=x$ to get the result. – mrs Jul 9 '13 at 5:57