Flipping a Switch: One-Sided Limits

The moment a switch closes, the current through an inductor can't jump, but the voltage across it can. Approach the switching time from the left and from the right and you get two different values. One-sided limits describe exactly that jump.

How it works

One-sided limits and continuity

In one sentence

$\lim_{x\to a^-}f(x) = L$ uses only $x < a$ and $\lim_{x\to a^+}f(x) = M$ only $x > a$ ; $\lim_{x\to a}f(x)$ exists exactly when both exist and are equal, and $f$ is continuous at $a$ when that limit equals $f(a)$ .

General formula

\lim_{{\htmlData{sym=1}{t\to0^-}}} {\htmlData{sym=0}{v(t)}} = 0, \qquad \lim_{{\htmlData{sym=2}{t\to0^+}}} {\htmlData{sym=0}{v(t)}} = {\htmlData{sym=3}{V_s}}, \qquad \lim_{t\to0} {\htmlData{sym=0}{v(t)}} \text{ does not exist}

Walk up to the switching time from the past and you see one value; walk back to it from the future and you see another. A function is continuous only where both walks agree.

What the symbols mean

$v(t)$
Inductor voltage
Jumps when the switch closes.
$t\to0^-$
From the left
Approaching the switching time from before it.
$t\to0^+$
From the right
Approaching from after it.
$V_s$
Supply voltage
The voltage that appears the instant the switch closes.

Before $t = 0$ the switch is open and the voltage is 0; after, it is $V_se^{-t/\tau}$ .
The left-hand limit $t \to 0^-$ uses only times before the switch: it is 0.
The right-hand limit $t \to 0^+$ uses only times after: it is $V_s$ .
They differ, so the two-sided limit doesn't exist and $v$ is discontinuous at 0. The current, by contrast, has equal one-sided limits: it is continuous.

Learn the math

One-Sided LimitsCalc I · Lesson 6, Limits ContinuityCalc I · Lesson 12, Limits