Consider a linear system
 $$Ax=b\qquad (\*)$$
 and a sequence of perturbed linear systems $$(A+\delta A\_n)x=b+\delta b\_n. \qquad (n)$$
 Suppose that all the linear systems are consistent (i.e., have solutions). 


My question is: Let $\overline{x}$ be a solution to $(\*)$. Suppose that $\delta A\_n \rightarrow 0$ and $\delta b\_n \rightarrow 0$. Does there exist a sequence of solutions $\overline{x}\_n$ such that each $\overline{x}\_n$ is a solution to the linear system $(n)$, satisfying $\overline{x}\_n \rightarrow \overline{x}$?


A possible helpful result is that for any solution $x\_n$ to the linear system $(n)$, there exists a solution $x$ to the linear system $(\*)$ such that
 $$\frac{\|x\_n - x\|}{\|x\|}\leq \frac{\|A^\sharp\|\|A\|}{1-\|A^\sharp\|\|\delta A\_n\|}\left(\frac{\|\delta b\_n\|}{\|b\|}+\frac{\|\delta A\_n\|}{\|A\|}\right)\quad \text{if} \ \|A^\sharp\| \| \delta A\_n\| <1,$$
 where $A^\sharp$ denotes the (weighted) Moore-Penrose inverse or the Drazin inverse.
 However, the existence of a solution in this result is for the linear system $(\*)$ rather than the linear system $(n)$. The above result also holds if we replace $(x, x\_n, A)$ with $(x\_n, x, A+\delta A\_n)$ respectively by considering the $( \* )$ as a perturbation of $(n)$. However, if so, 
$x\_n$ and $(A+\delta A\_n)^\sharp$ are not guaranteed to be bounded and thus we cannot have a upper bound of $\|x\_n-x\|$ tending to $0$. (Correct?) 


Anyone can help? Thank you!



A question for solutions of perturbed linear systems
Consider $A= (1, 0 ; 1,0)$ and $b= (1 ; 1)$.
Let $x$ be any solution of this, so $x= (1; t)$ for some $t$.


Now, perturbe this to $(1, 0 ; 1 - 1/n,1/n^2)$ and $b$ unchanged, so pertruebed by $0$ (but one could also impose some nontrivial perturbation). 
The perturbation(s) tends to zero. But the (only) solution is $(1; n)$. 
Thus $x\_n$ does not converge, and one seees that the $x\_n$ and the inverse (in this case it is actually invertible, but one could expand the system to avoid this) of the perturbed system are not bounded. 


