Proving that $I-EA^{-1} = I+EA^{-1} + o(RelError(\tilde{A},A))$ Let $A\in\mathbb{R}^{n\times n}$ be a non-singular matrix and let $\tilde{A} = A-E$ be an approximation of $A$. The relative error of this approximation is $$RelError(\tilde{A},A) = \frac{\| \tilde{A}-A\|}{\|A\|} = \frac{\|E\|}{\|A\|}.$$ In my book, they say that $$(I-EA^{-1})^{-1} = I + EA^{-1} + o(RelError(\tilde{A},A)).$$ I know that $(I-EA^{-1})^{-1} = I + EA^{-1} + (EA^{-1})^2 + (EA^{-1})^3+\ldots$, so this means $(EA^{-1})^2 + (EA^{-1})^3+\ldots = o(RelError(\tilde{A},A)) = o\Big(\frac{\|E\|}{\|A\|}\Big)$. Using the definition of the little oh, I just need to prove that $$\lim_{E\to 0} \frac{(EA^{-1})^2 + (EA^{-1})^3+\ldots}{\frac{\|E\|}{\|A\|}} = \|A\|\cdot \lim_{E\to 0} \frac{(EA^{-1})^2 + (EA^{-1})^3+\ldots}{\|E\|} = 0.$$ To be honest, I never worked with a limit of matrices, not sure how to proceed from here. I hope you can help me here, thanks.

I'm not sure but maybe you can use this: $$\|A\|\lim_{\|E\|\to 0}\|\frac{(EA^{-1})^2 + (EA^{-1})^3+\ldots}{\|E\|}\|\le\|A\|\lim_{\|E\|\to 0}\|\frac{\|(EA^{-1})^2\| + \|(EA^{-1})^3\|+\ldots}{\|E\|}\|=$$ $$=\|A\|\|A^{-1}\|^2\lim_{\|E\|\to 0}\|\frac{\|E\|^2 + \|E\|^3\|(A^{-1})\|+\ldots}{\|E\|}\|=$$ $$=\|A\|\|A^{-1}\|^2\lim_{\|E\|\to 0}({\|E\| + \|E\|^2\|(A^{-1})\|+\ldots})$$ If we want this equation $$(I-EA^{-1})^{-1} = I + EA^{-1} + (EA^{-1})^2 + (EA^{-1})^3+\ldots$$ we need $\|EA^{-1}\|<1$. That means $\|EA^{-1}\| + \|(EA^{-1})\|^2 + \|(EA^{-1})\|^3+\ldots$ converges to $\gamma$. And now we have $$\|A\|\|A^{-1}\|^2\lim_{\|E\|\to 0}({\|E\| + \|E\|^2\|(A^{-1})\|+\ldots})=\|A\|\|A^{-1}\|^2\lim_{\|E\|\to 0}({\|E\| + \|E\|*\gamma})=0$$