Prove $$\lim_{\overline{h}\rightarrow\overline{0}}||\frac{(h_1+h_2)^2sin(\frac{\pi}{h_1+h2})-h_1\pi-h_2\pi}{h}||=0$$
My work: $$\lim_{\overline{h}\rightarrow\overline{0}}||\frac{(h_1+h_2)^2\sin(\frac{\pi}{h_1+h2})-h_1\pi-h_2\pi}{h}||=\lim_{\overline{h}\rightarrow\overline{0}}\frac{||(h_1+h_2)^2\sin(\frac{\pi}{h_1+h2})-h_1\pi-h_2\pi||\pi}{\sqrt{h_1^2+h_2^2}\,||\frac{\pi}{h_1+h_2}||\,||h_1^2+h_2^2||}\leq \lim_{\overline{h}\rightarrow\overline{0}}\frac{||(h_1+h_2)^2\sin(\frac{\pi}{h_1+h_2})||}{\sqrt{h_1^2+h_2^2}\,||\frac{\pi}{h_1+h_2}h_1^2+h_2^2||}+\lim_{\overline{h}\rightarrow \overline{0}}\frac{||h_1\pi+h_2\pi||}{\sqrt{h_1^2+h_2^2}\,||\frac{\pi}{h_1+h_2}h_1^2+h_2^2||}=\lim_{\overline{h}\rightarrow\overline{0}}\frac{||(h_1+h_2)^2||}{h_1^2+h_2^2}+\lim_{\overline{h}\rightarrow\overline{0}}\frac{||h_1\pi+h_2\pi||}{\sqrt{h_1^2+h_2^2}\,||\frac{\pi}{h_1+h_2}h_1^2+h_2^2||}$$
here I'm stuck, can someone help me?
$$S=lim_{\overline{h}\rightarrow\overline{0}}||\frac{(h_1+h_2)^2sin(\frac{\pi}{h_1+h2})-h_1\pi-h_2\pi}{h}||=lim_{\overline{h}\rightarrow\overline{0}}||\frac{(h_1+h_2)^2sin(\frac{\pi}{h_1+h2})-h_1\pi-h_2\pi}{\sqrt{h_1^2+h_2^2}}||=lim_{\overline{h}\rightarrow\overline{0}}||\frac{(h_1+h_2)^2sin(\frac{\pi}{h_1+h2})-h_1\pi-h_2\pi}{h_1+h_2}||\lim_{\overline{h}\rightarrow\overline{0}}\frac{h_1+h_2}{\sqrt{h_1^2+h_2^2}}$$since $\overline{h}\rightarrow\overline{0}$ so is that $h_1+h_2\to 0$ therefore by substituting $u=h_1+h_2$ we obtain: $$S=\lim_{u\to 0}\frac{u^2\sin {\pi\over u}-\pi u}{u}\lim_{\overline{h}\rightarrow\overline{0}}\frac{h_1+h_2}{\sqrt{h_1^2+h_2^2}}=(\lim_{u\to 0}u\sin {\pi\over u}-\pi )\lim_{\overline{h}\rightarrow\overline{0}}\frac{h_1+h_2}{\sqrt{h_1^2+h_2^2}}$$$$=-\pi\lim_{\overline{h}\rightarrow\overline{0}}\frac{h_1+h_2}{\sqrt{h_1^2+h_2^2}}$$which doesn't converge at all.
Another way to show the nonexistence of the limit is to consider two different paths $h_2=0,h_1\to 0$ and $h_1=h_2\to 0$. The former one leads to:$$S=lim_{h_1\to 0}|\frac{h_1^2\sin \frac{\pi}{h_1}-\pi h_1}{h_1}|=lim_{h_1\to 0}|{h_1\sin \frac{\pi}{h_1}-\pi \frac{h_1}{|h_1|}}|=\pi$$and the lattes one leads to $S=2\pi$ and implies what we want.