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I am trying to understand an article mentioned here https://www.probabilitycourse.com/chapter4/4_3_2_delta_function.php. Some where down the explanation there's a statement marked (4.9) and it says

$u(x) = \lim\limits_{\alpha \to 0} u_\alpha(x)$ , $\alpha > 0$

Where $u(x)$ is defined to be a step function

\begin{equation} \hspace{50pt} u(x) = \left\{ \begin{array}{l l} 1 & \quad x \geq 0 \\ 0 & \quad \text{otherwise} \end{array}\right\} \hspace{50pt} \end{equation} And $u_\alpha(x)$ is defined to be as below \begin{equation} \nonumber u_{\alpha}(x) = \left\{ \begin{array}{l l} 1 & \quad x > \frac{\alpha}{2} \\ \frac{1}{\alpha} (x+\frac{\alpha}{2}) & \quad -\frac{\alpha}{2} \leq x \leq \frac{\alpha}{2} \\ 0 & \quad x < -\frac{\alpha}{2} \end{array} \right\} \end{equation} I tried taking the limit of $u_\alpha(x)$ as $\alpha$ approaches zero but i am not able to arrive at this statement $u(x) = \lim\limits_{\alpha \to 0} u_\alpha(x)$

Shouldnt the limit of $\lim\limits_{\alpha \to 0} \frac{1}{\alpha} (x+\frac{\alpha}{2})$ when $-\frac{\alpha}{2} \leq x \leq \frac{\alpha}{2} $ be equal to $\infty$ ?.

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  • $\begingroup$ If $0<\alpha < x/2$, then $u_{\alpha}(x)=1$. $\endgroup$ – enedil Mar 14 at 18:17
  • $\begingroup$ First observe that as $\alpha \to 0,$ the interval you're interested in shrinks to the point $x=0.$ That might be significant. But perhaps they made some other assumption you've not spotted. Meh. $\endgroup$ – Allawonder Mar 14 at 18:41
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Observe that as $\alpha \to 0,$ the interval $[-\alpha/2,\alpha/2]$ shrinks to the point $\{0\},$ and $(-\infty, -\alpha/2)\cup (\alpha/2,+\infty)$ goes to $(-\infty,+\infty),$ so that the result follows immediately.

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