Laplace Transforms

[pazzy78]

Curious where he got the 0 from ?

is e^infinity = 0 ?

 

[BigBeachBanana]

is e^infinity = 0 ?

Negative infinity.

 

[pazzy78]

Yes but - infinity still not 0

 

[BigBeachBanana]

Yes but - infinity still not 0

[math]\lim_{x\to -\infty} e^x = 0[/math]or
[math]\lim_{x\to \infty} e^{-x} = 0[/math]

 

[pazzy78]

OK im a little rusty on my calculas, isn't the limit of something the same as differentiation ?

Whereas this is integration and plugging limits...

 

[mario99]

OK im a little rusty on my calculas, isn't the limit of something the same as differentiation ?

Whereas this is integration and plugging limits...

You're half right.

Every differentiation is a limit, but not every limit is a differentiation. When we have an improper integral (a definite integral that has either or both limits infinite), we use the limit as infinite is not a number that we can plug in. However, you can violate the rule and plug infinite directly, but your intention must be just a shortcut of using the limit.

The shortcut is:

[imath]\displaystyle e^{-\infty} = 0[/imath]

instead of

[imath]\displaystyle \lim_{x\rightarrow -\infty}e^{x} = 0[/imath]

 

[blamocur]

OK im a little rusty on my calculas, isn't the limit of something the same as differentiation ?

Whereas this is integration and plugging limits...

Limits are in the formal definitions of both definite integrals and derivatives. But more to the point:
[math]\int_a^\infty f(x) dx = \lim_{b\rightarrow\infty} \int_a^b f(x) dx[/math]