For any function
𝑓
(
𝑡
)
whose derivative is piecewise continuous and of exponential order
𝛼
on

[
0
,
∞
)
,
𝐿
{
𝑓
′
}
(
𝑠
)
=
lim
𝑠
⟶
∞
𝑠
𝐿
{
𝑓
}
(
𝑠
)
−
𝑓
(
0
)
.

Question

For any function 
𝑓
(
𝑡
)
 whose derivative is piecewise continuous and of exponential order 
𝛼
 on

[
0
,
∞
)
, 
𝐿
{
𝑓
′
}
(
𝑠
)
=
lim
𝑠
⟶
∞
𝑠
𝐿
{
𝑓
}
(
𝑠
)
−
𝑓
(
0
)
.

BlackTom AI · Accepted Answer

The statement concerns a standard property of the Laplace transform for a function f whose derivative is piecewise continuous and of exponential order on [0, ∞).

First, recall the correct fundamental relation: if F(s) = L{f}(s) and f is of appropriate growth, then L{f′}(s) = s F(s) − f(0) for values of s where the transform exists. This is a direct result of integrating by parts and the definition of the Laplace transform.

Option evaluation: The given expression claims L{f′}(s) equals lim_{s→∞} [ s L{f}(s) − f(0) ].
- The correct formula does not involve a limit as s→∞. It is valid for each s (in the region of convergence) as L{f′}(s) = s F(s) − f(0). Treating the right-hand side as a limit ignores the dependence on s and conflates a pointwise identity with a limiting process.
- Moreover, taking a limit as s→∞ of s F(s) − f(0) is not generally equal to L{f′}(s) for large s, since L{f′}(s) is a function of s, not a limit that collapses to a single number.

Why the statement is false: the absence of the explicit s-dependence on the left side (L{f′}(s)) in the proposed limit form creates a mismatch. The standard relation L{f′}(s) = s F(s) − f(0) holds for every s in the domain of convergence; turning it into a limit as s→∞ changes the meaning entirely and can yield an incorrect value.

Why the conditions matter: the hypotheses about piecewise continuity and exponential order guarantee that L{f} exists (so F(s) is well-defined) and that L{f′} can be expressed via the standard formula, but they do not justify replacing the formula with a limit expression. The limit form simply does not reflect the true, pointwise relation between L{f′} and F.

In summary, the statement misstates the Laplace transform property by introducing a limit where a direct, s-dependent equality should hold. The correct statement is L{f′}(s) = s F(s) − f(0), not L{f′}(s) = lim_{s→∞} [ s F(s) − f(0) ].

MAP2302 L19 section 7.3

For any function 𝑓 ( 𝑡 ) whose derivative is piecewise continuous and of exponential order 𝛼 on [ 0 , ∞ ) , 𝐿 { 𝑓 ′ } ( 𝑠 ) = lim 𝑠 ⟶ ∞ 𝑠 𝐿 { 𝑓 } ( 𝑠 ) − 𝑓 ( 0 ) .

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