Q:

Does a small perturbation push the minimizing point of continuous function too far?

Let $(M,\,d)$ be a compact metric space and $f,\,g: M \rightarrow \mathbb{R}$ two continuous functions. 
For $|ε|<1$ we define a perturbation of $g$: $$g_ε(x) = g(x) - εf(x)$$
For every $ε>0$, there exist some $x_ε\in M$ such that $\displaystyle g_ε(x_ε) = \min_{x\in M}g_ε(x)$, since $g_ε$ is continuous and $M$ is compact. 
The same argument applies for $g$, too, but suppose that the according minimizer $x_0$ is unique.
Can we argue that for small $ε$ we can find a $x_ε$ $δ$-close to $x_0$ for $δ\xrightarrow{ε\rightarrow 0} 0$? 
How can we get around it?
Every help is appreciated! Thank you in advance!

A:

It is straightforward to show that any accumulation point taken from $\operatorname{argmin}  g_{1 \over n}$ will be an element of $\operatorname{argmin}  g$:
Suppose $g_{1 \over n}( x_{1 \over n}) \le g_{1 \over n}(x)$ for all $n,$ and $x \in M$. Suppose $x^*$ is an accumulation point of $x_{1 \over n}$, then taking limits along an appropriate subsequence gives: $g(x^*) \le g(x)$ for all $x$.
Hence if $\operatorname{argmin}  g$ is a singleton, any sequence taken from $\operatorname{argmin}  g_{1 \over n}$ must converge to this point. To see this note that any subsequence of $x_{1 \over n}$ must contain a further subsequence that converges to $\operatorname{argmin}  g$. Hence $x_{1 \over n}$ converges.

