Prove that $f(t,x)=x^{1/2}$ defined on a rectangle $\mathrm{R}=\{(t,x):|t|\le 2,\;\; |x|\le 2\}$ does not admit partial derivates with respect to $x$ at $(0,0)$
My Solution :
$f(t,x)=x^{1/2}$ is defined on a rectangle $\mathrm{R}=\{(t,x):|t| \le 2 \;\; |x| \le 2\}$
Let's first prove that $f(t,x)$ is unbounded:
$$|f(t,x)-f(t,y)| = |x^{1/2}-y^{1/2}| \le L|x-y| \rightarrow \dfrac{|x^{1/2}-y^{1/2}|}{|x-y|} \le L$$
But if we choose x and y so close to zero than L will be infinity. That is $f(t,x)$ does not satisfy Lipschitz Condition. Now, here is my question:
$$f_{x}(0,0) = \displaystyle\lim_{h \to 0} \dfrac{f(0,h)-f(0,0)}{h} = \displaystyle\lim_{h \to 0} \dfrac{h^{1/2}}{h} =\displaystyle\lim_{h \to 0} \dfrac{1}{h^{1/2}}=\infty$$
Is it true? Thanks in advance!