# Thread: Limit question as part of a Laplace Transform

1. ## Limit question as part of a Laplace Transform

In trying to get the Laplace Transform of

$f(t) = te^{at}$

...I end up at this step close to the end:

$\lim_{A \to \infty} \left[ \frac{A}{a - s} \; \frac{e^{aA}}{e^{sA}} - \frac{1}{(a - s)^2} \; \frac{e^{aA}}{e^{sA}} + \frac{1}{(a - s)^2} \right]$

I know the answer to this is

$\frac{1}{(a - s)^2}$

...which means the first two sets of terms equal zero. But I'm not sure I understand why. At best, the exponential fractions seem like they should converge to 1, so I'm not sure how to get to zero.

Can someone help me understand this?

2. ## Re: Limit question as part of a Laplace Transform

Originally Posted by Lancet
In trying to get the Laplace Transform of

$f(t) = te^{at}$

...I end up at this step close to the end:

$\lim_{A \to \infty} \left[ \frac{A}{a - s} \; \frac{e^{aA}}{e^{sA}} - \frac{1}{(a - s)^2} \; \frac{e^{aA}}{e^{sA}} + \frac{1}{(a - s)^2} \right]$

I know the answer to this is

$\frac{1}{(a - s)^2}$

...which means the first two sets of terms equal zero. But I'm not sure I understand why. At best, the exponential fractions seem like they should converge to 1, so I'm not sure how to get to zero.

Can someone help me understand this?
If $s> a\ge 0$ the exponential $e^{A(a-s)}$ goes to zero as $A$ goes to infinity.

CB

3. ## Re: Limit question as part of a Laplace Transform

Originally Posted by CaptainBlack
If $s> a\ge 0$ the exponential $e^{A(a-s)}$ goes to zero as $A$ goes to infinity.

CB
Okay, that I understand, but how can we make that assumption that $s > a >= 0$ ?

4. ## Re: Limit question as part of a Laplace Transform

Originally Posted by Lancet
Okay, that I understand, but how can we make that assumption that $s > a >= 0$ ?
If not, then the Laplace transform does not exist.

5. ## Re: Limit question as part of a Laplace Transform

Originally Posted by mr fantastic
If not, then the Laplace transform does not exist.
Oh, I see. That's the only possible scenario where that limit converges, so we go with that and set the condition that makes it true.

Thank you both!