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I'm working on Laplace Transforms and have got to a section where they are talking about zero to the power plus or minus and that they are different. I can't remember what this means though.

It's generally used in limits.

$\lim\limits_{t\to 0^-}$ or $\lim\limits_{t\to 0^+}$

Any help would be much appreciated.

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This is not a power or exponent. It is part of the notation for limits. – Code-Guru Aug 7 '12 at 13:53
up vote 9 down vote accepted

Say we let

$$H(x)=\begin{cases} 0, & x < 0, \\ 1, & x > 0, \end{cases}$$

and let $H(0)$ be not defined.

Say I would like to approach $0$ on this function. However, a problem arises! Looking at the plot of the function, it is clear that if one were to approach from the right hand side, the limit is $1$, whilst if one approaches from the left, the limit is $0$ and thus the two-sided limit does not exist (both sides should be approaching the same number for this limit to exist)! This can also be easily seen by plugging in numbers:

$$H(1)=1$$ $$H(.1)=1$$ $$H(.000000000001)=1$$ etc. But, doing the same thing from the left hand side, we find

$$H(-1)=0$$ $$H(-.1)=0$$ $$H-(.000000000001)=0$$

Step function

Thus we need to define a different type of limit for functions with similar discontinuities so we may approach from either side. This limit is the "one-sided limit" and is used generally when a two-sided limit does not exist, like in the above case. $\lim_{x \to x_0^+}f(x)$ represents the right handed limit of $f(x)$ to $x_0$ whilst $\lim_{x \to x_0^-}f(x)$ represents the left hand limit. So we see that $\lim_{x \to 0} H(x)$ does not exist, but

$$\lim_{x \to 0^+}H(x)=1$$ $$\lim_{x \to 0^-}H(x)=0$$

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Nice example, especially with the plot. I think you mean for the value of $H(x)$ to be determined by the value of $x$ though, not $n$. – flodyninja Aug 7 '12 at 22:25
@flodyninja Ah, yes. Thanks for noticing! – Argon Aug 7 '12 at 22:59

In this case, the plus and minus refer to the direction from which you approach zero. So,

$\lim \limits_{t \to 0^{-}}$

means the limit as $t$ approaches $0$ from the negative side, or from below, while

$\lim \limits_{t \to 0^{+}}$

means the limit as $t$ approaches $0$ from the possitive side, or from above. So, it is just specifying which direction you are moving along the number line.

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your answer doesnt have any link to Laplace Transform. – Seyhmus Güngören Aug 7 '12 at 14:37
The question was regarding notation, the context that the notation was found in was the use of Laplace transforms. The notation is not unique to that one application, so there was no need to mention Laplace transforms in my answer. – flodyninja Aug 7 '12 at 22:15

And a good, visual example could be something like

                            enter image description here

$$\begin{align} \lim_{x\to x_0^-}f&=y_1\\ \lim_{x\to x_0^+}f&=y_2 \end{align}$$

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Good example, but I prefer $f(x)=\frac{1}{x}$. You don't even need to graph it to see the difference between the two limits. – emory Aug 7 '12 at 16:45
@emory, "You don't even need to graph it..." - you and me, maybe. Some kids these days, on the other hand... – J. M. Aug 7 '12 at 23:46

$$\lim_{t \to 0^{+}}$$ indicates that the limit is meant to be taken only from the positive direction; it's a one-sided limit

Right hand Limit: $\displaystyle\lim_{t \to 0+} f(x) = \displaystyle\lim_{t \to 0} f(x+|t|) $.

Left Hand Limit: $\displaystyle\lim_{t \to 0-} f(x)= \displaystyle\lim_{t \to 0} f(x-|t|) $

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The notation here is confusing: there are limits over $t$ of functions which don't involve $t$. Also $\lim_{t\to 0}f(x+t) = \lim_{t\to 0}f(x-t)$; these are just the usual two-sided limits. Perhaps you mean something like $\lim_{t\to 0^+} f(x+t) = lim_{s\to+\infty} f\left(x+\frac{1}{s}\right)$ and $\lim_{t\to 0^-} f(x+t) = lim_{s\to-\infty} f\left(x+\frac{1}{s}\right)$? – Noah Stein Aug 7 '12 at 14:52
$$\lim_{x \to 0}f(x)=\lim_{x \to \infty}f\left(\frac{1}{x}\right)$$ – Argon Aug 7 '12 at 15:45
Another alternative is to write $\displaystyle \lim_{x \to 0^+} f(x) = \lim_{x \to 0} f(|x|)$ and $\displaystyle \lim_{x \to 0^-} f(x) = \lim_{x \to 0} f(-|x|)$. – Clive Newstead Aug 7 '12 at 23:25

protected by Asaf Karagila Oct 9 '14 at 15:37

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