Find the relative condition number of $f$. $$f(x_1, x_2)= \frac{x_1}{x_2}$$
So, when I use the definition of the relative condition number $\kappa$, I get:
$$\kappa(f,x)= \lim_{\epsilon \rightarrow 0^{+}} \sup_{\|\delta x\| \leq \epsilon} \frac{\|f(x+ \delta x) - f(x) \| \cdot \|x\|}{\|\delta x\| \cdot \|f(x)\|} \\ = \lim_{\epsilon \rightarrow 0^{+}} \sup_{\|\delta x\| \leq \epsilon} \frac{\|\frac{x_1 + \delta x_1}{x_2 + \delta x_2} - \frac{x_1}{x_2} \| \cdot \|x\|}{|\delta| \| x\| \cdot \|\frac{x_1}{x_2}\|} \\ =\lim_{\epsilon \rightarrow 0^{+}} \sup_{\|\delta x\| \leq \epsilon} \frac{\|\frac{x_1 + \delta x_1}{x_2 + \delta x_2} - \frac{x_1 + \delta x_1}{x_2 + \delta x_2} \|}{|\delta| \cdot \|\frac{x_1}{x_2}\|} \\ =\lim_{\epsilon \rightarrow 0^{+}} \sup_{\|\delta x\| \leq \epsilon} \frac{\|\frac{0}{x_2 + \delta x_2}\|}{|\delta| \cdot \|\frac{x_1}{x_2}\|} \\ =\lim_{\epsilon \rightarrow 0^{+}} \sup_{\|\delta x\| \leq \epsilon} \frac{0}{|\delta| \cdot \|\frac{x_1}{x_2}\|} \rightarrow \infty $$
I don't think this can be right. If anyone could review my calculations, it would be great. Thank you.