I know in 1-dimension case, $$f(x)=f(x_0)+f'(\tilde{x_0})(x-x_0)$$ where $\tilde{x_0}$ lies between $x$ and $x_0$.
What about in k-dimension case? $$\mathbf{f}(\mathbf{x})=\mathbf{f}(\mathbf{x}_0)+\mathbf{f}'(\tilde{\mathbf{x}_0})(\mathbf{x}-\mathbf{x}_0)$$ Does $\mathbf{\tilde{x}}_0$ still lies between $\mathbf{x}$ and $\mathbf{x}_0$?
If $\mathbf{x}\rightarrow\mathbf{x}_0$, does $\mathbf{\tilde{x}}_0\rightarrow\mathbf{x}_0$?
Consider $\mathbb C$ which can be treated as $\mathbb R^{2}$. Consider teh function $f(z)=e^{z}$. $e^{0}-e^{2\pi i}=0$ cannot be written as $2\pi i e^{z}$ for any $z$.