# Average of left and right limits | Signum function, Heaviside step function, and Grandi's series

This question probably already has an answer but usually involves stuff that's way over the top of my head so I'm hoping for a simple explanation.

In Adams, R. A., & Essex, C. (7th edition) Calculus: A Complete Course (p. 67, ex. 6):

$$\lim_{x \to 0-} sgn(x) = -1 \hskip 3em \text{and} \hskip 3em \lim_{x \to 0+} sgn(x) = 1$$ Since these left and right limits are not equal, $lim_{x \to 0}\ sgn(x)$ does not exist.

Applying this same rule to the Heaviside step function, $$H[n] = \left\{\begin{array}{11}0, & \quad n < 0, \\ 1, & \quad n \geq 0, \end{array}\right.$$ the $lim_{x \to 0}\ H(x)$ should also not exist. However I've seen Grandi's series as: $$\sum_{n=0}^{\infty}(-1)^n = 1/2$$

From my point of view these two problems feel almost the same, except the unit step function is being evaluated at 0 while Grandi's series is at infinity. Are they even comparable? And if so, does Grandi's series merely overstep this rule because it doesn't make sense to approach positive infinity from the right? I'd really like to know.

• Grandi's series is divergent. In $\sum_{n = 0}^{\infty} (-1)^n = 1/2$, the $\sum_{n = 0}^{\infty}$ has a different meaning than the ordinary one. If you use a custom meaning of $\lim$, you can also say $\lim_{x\to 0} H(x) = 1/2$. – Daniel Fischer May 12 '16 at 20:13
• If the standard definition of $\lim$ requires the two-sided limits to be equal then how is it consistent with infinity? @DanielFischer – Pae loah May 12 '16 at 21:55
• It's better not to focus on "two-sided", that only works in dimension $1$, and if the function is defined on both sides of the point under consideration. If a function is defined on $[a,b]$, or on $(a,b)$, then the limit at $a$ is a one-sided limit. And if a function is defined on a higher-dimensional domain, e.g. $\{(x,y)\in\mathbb{R}^2:0<x^2+y^2< 1\}$, then the concept of one-sided limit isn't appropriate. For the case of functions defined on subsets of $\mathbb{R}$, one can however say that the function is only defined on one side of $\pm\infty$, so a limit can't be other than one-sided. – Daniel Fischer May 13 '16 at 8:35