Slowly-varying envelope approximation: what does it imply?

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.

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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 $V(x)$.