Let $\alpha_\delta >0$ be a quantity that depends only on $\delta$ and let $I_\varepsilon$ be defined as follows: $$I_\varepsilon = \alpha_\delta + \beta_{\varepsilon, \delta}$$ where $\beta_{\varepsilon, \delta}$ depends both on $\varepsilon$ and on $\delta$.
I've got stuck on the following passage which is part of a demostration I'm studying on some class notes: $${\max\lim}_{\varepsilon \to 0}{|I_\varepsilon|} \leq {\max \lim}_{\varepsilon\to 0}\{\alpha_\delta + |\beta_{\varepsilon, \delta}|\} \leq (?) \alpha_\delta + \max\lim_{\varepsilon\to 0}|\beta_{\varepsilon,\delta}|:$$ the 2nd "$\leq$" makes sense, but I can't understand the reason why $\quad{\max \lim}_{\varepsilon\to 0}\{\alpha_\delta + |\beta_{\varepsilon, \delta}|\}\quad$ is not equal to $\quad \alpha_\delta + \max\lim_{\varepsilon\to 0}|\beta_{\varepsilon,\delta}|\quad$ if $\alpha_\delta$ doesn't depend on $\varepsilon$.
Can you help me understand it?
Thank you.
I assume you already got this first bit:
\begin{align}\max\lim_{\varepsilon\to 0}|I_\varepsilon|&=\max\lim_{\varepsilon\to 0}(|\alpha_\delta+\beta_{\varepsilon,\delta}|)\\ &\le\max\lim_{\varepsilon\to 0}(\alpha_\delta+|\beta_{\varepsilon,\delta}|)\\ &=\max\big(\lim_{\varepsilon\to 0}(\alpha_\delta)+\lim_{\varepsilon\to 0}|\beta_{\varepsilon,\delta}|\big)\\ &=\max\big(\alpha_\delta+\lim_{\varepsilon\to 0}|\beta_{\varepsilon,\delta}|\big).\end{align}
Now the key point is that for any $f,g\colon\Bbb{R}\to\Bbb{R}_{\ge 0}$, $$\max_x(f(x),g(x))\le \max_x(f(x))+\max_x(g(x)).$$ They are not necessarily equal (consider when $f=g$, for example).
In your case you have two positive "functions" of $\delta$ (namely, $\alpha_\delta$ and $\lim_{\varepsilon\to 0}|\beta_{\varepsilon,\delta}|$), and hence $$\max\big(\alpha_\delta+\lim_{\varepsilon\to 0}|\beta_{\varepsilon,\delta}|\big)\le\max\alpha_\delta+\max\lim_{\varepsilon\to 0}|\beta_{\varepsilon,\delta}|.$$