Slowly-varying envelope approximation: what does it imply? I understand that the slowly-varying envelope approximation means that we can write an electromagnetic wave as
$$ E(x,t)=V(x,t)e^{i(k_0x-\omega_0 t)},$$
where
$$ \left \vert \frac{dV}{dx} \right \vert \ll |k_0 V(x,t)| \quad \text{and} \quad \left \lvert \frac{dV}{dt} \right \rvert \ll |\omega_0 V(x,t)| \, .$$
I have two questions about this approximation:


*

*Why this implies that the band of the wave is narrow?

*What is the maximum value of $Δx$, compared to $\lambda$, for which $V(x+Δx)≈V(x)$ still holds?

 A: 1. Narrow-band-ness
Consider, as JEB's answer does, the pure-space dependence
$$ E(x,0)=V(x,0)e^{ik_0x},$$
and let's look at the Fourier transform of this (dropping the $t=0$'s for simplicity):
\begin{align}
\tilde E(k)
& = 
\int E(x)e^{-ikx}\mathrm dx 
= 
\int V(x)e^{-i(k-k_0)x}\mathrm dx
= \tilde{V}(k-k_0),
\end{align}
i.e. the frequency-space relationship is, as noted by JEB, the convolution of $\tilde V(k)$ and the Dirac delta that results from Fourier transforming $e^{ik_0x}$. Let's look at $\tilde V(k)$ more closely, though, by first considering the Fourier transform of its derivative $\frac{\mathrm dV}{\mathrm dx}$:
\begin{align}
\int \frac{\mathrm dV}{\mathrm dx}(x) e^{-ikx}\mathrm dx 
& =
-ik \int V(x) e^{-ikx}\mathrm dx
=-ik\tilde V(k),
\end{align}
assuming that the boundary terms in $V(x)|_{-\infty}^\infty$ vanish. This means, therefore, that
\begin{align}
|k\tilde V(k)|
& = 
\left|
\int \frac{\mathrm dV}{\mathrm dx}(x) e^{-ikx}\mathrm dx 
\right|
\\ & \ll
 \left|
\int k_0 V(x) e^{-ikx}\mathrm dx 
\right|
 = 
k_0|\tilde V(k)|
.
\end{align}
Taken at face value, that's one heck of a strange statement, but the way to read it is this: within the band in which $|\tilde V(k)|$ is appreciable, you must have $|k|\ll k_0$.
When applied to the full electric field, this then tells you that within the band in which $|\tilde E(k_0 + \Delta k)|$ is appreciable, you must have $|\Delta k| \ll k_0$, and this is the precise meaning that we assign to $E(x)$ being a narrowband wavepacket.

2. $V(x+Δx)≈V(x)$
Your second question,

[given that $|\frac{dV}{dx}|\ll |k_0 V(x,t)|$,] what is the maximum value of $Δx$, compared to $\lambda$, for which $V(x+Δx)≈V(x)$ still holds?

is rather harder, because neither of the statements involved, $|\frac{dV}{dx}|\ll |k_0 V(x,t)|$ and $V(x+Δx)≈V(x)$, is a hard quantitative statement.
Nevertheless, the way to tackle this is via a Taylor expansion,
\begin{align}
| V(x+\Delta x) - V(x) |
& =
|\Delta x \frac{\mathrm dV}{\mathrm dx}(x)| + \mathcal O(\Delta x^2)
\\ & \ll
|\Delta x \: k_0 V(x)| + \mathcal O(\Delta x^2)
\\ & =
2\pi \left|\frac{\Delta x}{\lambda} V(x)\right| + \mathcal O(\Delta x^2)
.
\end{align}
If you have a quantitative statement for what that "$\ll$" means, then it will directly give you a statement for the fractional change in terms of $\Delta x/\lambda$ displacements that you will observe in $V(x)$.
A: Let's look at fixed time:
$$ E(x) = V(x)e^{-ik_0x} \equiv V(x)\phi_{k_0}(x) $$
I wrote $\phi(x)$ to highlight the fact that it is a fast phase modulation of a slowly varying function.
What does slowly varying mean? It means the Fourier transform:
$$ \tilde V(k) = \int{e^{ikx}V(x)dx} $$
has its power below some $k_V$ that satisfies:
$$ ||k_V|| \ll k_0$$
It also has some bandwidth $\Delta k \approx 2 k_V$ (there a positive and negative components for a real $V(x)$).
All this condition means is that the wavelength of $V(x)$ is much longer than the carrier wavelength, $1/k_0$. How much depends on your system. If it's microwaves in a circuit, then you have to look at the hardware. If you are solving a boundary value problem, you have to look at the conditions of the problem. A factor of "100" is certainly good, "10" is OK, and "2" is pushing it.
As pointed out in the comments, the FT of a product is a convolution of the multiplicands' FTs, and:
$$ \tilde \phi(k) = \delta(k_0) $$
Now the convolution with a delta functions is a shift operation, hence:
$$ \tilde E(k) $$
looks like
$$ \tilde V(k) $$
shifted from 0 to that carrier frequency $k_0$.
Thus, the signal has some bandwidth $\Delta k$ centered around $k_0$.
Now you can revisit the "slow" question by saying $\Delta k/2 \ll k_0$--meaning the all the power is in the positive frequencies, which is the same as the earlier condition but view slightly differenlty. They are equivalent to the conditions on the derivative stated in the OP.
Adding the time-coordinate back in doesn't change anything, unless you have a difficult dispersion relation.
