# Understanding limit of a function in another article

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

$$$$\hspace{50pt} u(x) = \left\{ \begin{array}{l l} 1 & \quad x \geq 0 \\ 0 & \quad \text{otherwise} \end{array}\right\} \hspace{50pt}$$$$ And $$u_\alpha(x)$$ is defined to be as below $$$$\nonumber u_{\alpha}(x) = \left\{ \begin{array}{l l} 1 & \quad x > \frac{\alpha}{2} \\ \frac{1}{\alpha} (x+\frac{\alpha}{2}) & \quad -\frac{\alpha}{2} \leq x \leq \frac{\alpha}{2} \\ 0 & \quad x < -\frac{\alpha}{2} \end{array} \right\}$$$$ I tried taking the limit of $$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 $$\infty$$ ?.

• If $0<\alpha < x/2$, then $u_{\alpha}(x)=1$. – enedil Mar 14 at 18:17
• 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. – Allawonder Mar 14 at 18:41

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.