The following results let us evaluate some limits using the Laplace transform. This corresponds to:

Let’s get into the details!

## 1. Laplace Transform initial value

Proof:

Let us assume f is a bounded function. We may say there are a,K\in\mathbb{R} such that

|f(t)|\leq Ke^{at}

and consequently

e^{-st}|f(t)|\leq Ke^{-(s-a)t}

As F(s)=\mathscr{L}\{f(t)\}, applying the definition, we have

\displaystyle |F(s)|=\left|\int_{0}^{\infty}e^{-st}f(t)\right|dt\leq\int_{0}^{\infty}e^{-st}|f(t)|dt

And then

\displaystyle |F(s)|\leq\int_{0}^{\infty}e^{-(s-a)t}Kdt=K\int_{0}^{\infty}e^{-(s-a)t}dt

Since

\displaystyle\int_{0}^{\infty}e^{-(s-a)t}dt=\frac{1}{s-a}

we may conclude

\displaystyle |F(s)|\leq\frac{K}{s-a}

As

\displaystyle \lim_{s\to\infty}\frac{K}{s-a}=0

hence, from the squeeze theorem

\displaystyle \lim_{s\to\infty}F(s)=0

Now, let H(s) be Laplace Transform of f'(t):

H(s)=sF(s)-f(0)

If we assume f'(t) is bounded as the last result, then

\displaystyle\lim_{s\to\infty}H(s)=0

i.e.,

\begin{aligned} & \lim_{s\to0}(sF(s)-f(0))=0 \\ & \lim_{s\to0}sF(s)-f(0)=0 \\ & \lim_{s\to0}sF(s)=f(0) \end{aligned}

which concludes the proofing.

## 2. Illustrative example of initial value problem

Let us verify the theorem above for f(t)=\sin(t). As

\displaystyle F(s)=\frac{1}{s^{2}+1}

thus

\displaystyle sF(s)=\frac{s}{s^{2}+1}

Using the L’Hopital’s rule

\begin{aligned} \lim_{s\to\infty}sF(s) &=\lim_{s\to\infty}\frac{s}{s^{2}+1} \\ &=\lim_{s\to\infty}\frac{1}{2s} \\ &=0 \end{aligned}

Since

f(0)=\sin(0)=0

we may conclude

\displaystyle f(0)=\lim_{s\to\infty}sF(s)

## 3. Final value theorem Laplace

Proof:

We know from Laplace transform of derivatives:

\displaystyle\mathscr{L}\{f'(t)\}=\int_{0}^{\infty}e^{-st}f'(t)dt=sF(s)-f(0)

Taking the limit as s\to0

\begin{aligned} & \lim_{s\to0}\int_{0}^{\infty}e^{-st}f'(t)dt=\lim_{s\to0}[sF(s)-f(0)] \\ & \int_{0}^{\infty}\lim_{s\to0}e^{-st}f'(t)dt=\lim_{s\to0}sF(s)-f(0) \\ & \int_{0}^{\infty}1\cdot f'(t)dt=\lim_{s\to0}sF(s)-f(0) \\ & \lim_{t\to\infty}\int_{0}^{t}f'(t)dt=\lim_{s\to0}sF(s)-f(0) \end{aligned}

And from Fundamental Theorem of Calculus, the integral becomes

\begin{aligned} & \lim_{t\to\infty}\left.f(t)\right|_{t=0}^{t=t}=\lim_{s\to0}sF(s)-f(0) \\ & \lim_{t\to\infty}f(t)-f(0)=\lim_{s\to0}sF(s)-f(0) \end{aligned}

Hence

\displaystyle\lim_{t\to\infty}f(t)=\lim_{s\to0}sF(s)

## Illustrative Example of final value problem

If f(t) is a function in the variable t such that

\displaystyle F(s)=\mathscr{L}\{f(t)\}=\frac{3}{s+9}

then

\displaystyle sF(s)=\frac{3s}{s+9}

To evaluate \displaystyle\lim_{t\to\infty}f(t) is equivalent to solve the following limit

\begin{aligned} \lim_{s\to0}sF(s) & =\lim_{s\to0}\frac{3s}{s+9} \\ & =\frac{3\cdot0}{0+9} \\ & =\frac{0}{9} \\ & =0 \end{aligned}

Thus

\displaystyle\lim_{t\to\infty}f(t)=0

We reached the end of this lesson about asymptotic values and how they can be calculated using the Laplace transform. For a full list of Laplace transform properties, check this post!