Show that $f$ defined by $f(t,x)=|\sin(x)|+t$ satisfies a Lipschitz condition on the whole $tx-$ plane

Question

Show that $f$ defined by $f(t,x)=|\sin(x)|+t$ satisfies a Lipschitz condition on the whole $tx-$ plane with respect to its second argument, but $\dfrac{\partial f}{\partial x}$ does not existe when $x=0$. What fact does this ilustrate?

Answer 1

$|f(t,x)-f(s,y)|=|t-s+|\sin (x)| -|\sin (y)|| \leq |t-s|+| |\sin (x)| -|\sin (y)|| \leq |t-s|+ |\sin(x) -\sin (y)| \leq |t-s|+|x-y|$ by Mean Value Theorem. Hence $|f(t,x)-f(s,y)| \leq 2 ||((t,x) -(s,y)||$ where ||.|| is the usual norm on $\mathbb R^{2}$.

Answer 2

note that $$|\sin(x)+1|\le |\sin(x)|+1\le 1+1=2$$ i have used the triangle inequality $$|a+b|\le |a|+|b|$$ and that $$|\sin(x)|\le 1$$ for all real $x$ and $$|\sin(x)|$$ has no derivative at $$x=0$$ since $$\frac{f(t,0)-f(t,0)}{h}=\frac{|h|}{h}$$ doesn't exist for $h$ tends to zero