I am trying to understand the proof of the following theorem :

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> Let $A$ be an $n\times n$ matrix such that it has only negative eigenvalues and let $f: \mathbb{R}\times \mathbb{R}^n$ continuous and locally Lipschits in $x$. If $f(t,0)=0$ for every $t\in \mathbb{R}$ and $\lim\_{x\rightarrow 0}\frac{||f(t,x)||}{||x||}=0$ for any $t\in \mathbb{R}$ then the zero solution of the equation $x'=Ax+f(t,x)$ is asymptotically stable .
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Now in the proof of this, the authors claim that given $x\_0\in \mathbb{R}^n$ with $||x\_0||<\delta$ the solution satisfies $||x(t)||<\delta $ for any $t\in [t\_0,t\_1]$ sufficiently small. And then they use that fact to prove that if $||x\_0||<\frac{\delta}{M}$ then $||x(t)||<\delta$ using the following
\begin{align} 
x(t)&=e^{A(t-t\_0)}x\_0+\int\_{t\_0}^t e^{A(t-s)}f(s,x(s))ds
\\
\|x(t)\|&\leq Me^{-\alpha(t-t\_0)}\|x\_0\|+\int\_{t\_0}^tMe^{-\alpha(t-s)}\epsilon \|x(s)\|ds
\\
e^{\alpha t}\|x(t)\|&\leq Me^{\alpha t\_0}\|x\_0\|+\int\_{t\_0}^t M\epsilon e^{\alpha s}\|x(s)\|ds
\\
e^{\alpha t}\|x(t)\|&\leq Me^{\alpha t\_0}e^{M\epsilon (t-t\_0)}
\\[.5em]
\|x(t)\|&\leq Me^{(-\alpha +M\epsilon) (t-t\_0)}\|x\_0\|
\end{align}
I am just a bit confused why we are doing this for $||x\_0||<\frac{\delta}{M}$ since we know that if $||x\_0||<\delta$ then $||x(t)||<\delta$. I thought this might be an attempt to enlarge the interval where we have that $||x(t)||<\delta$, to use the fact that $|f(s,x(s)|<\epsilon||x(t)||$ we need the fact that $||x(t)||<\delta$ so we stay in the same interval.
Can anyone help me figure out why we care that $||x\_0||>\frac{\delta}{M}$? Thanks in advance.