Derivation for wavelength bandwidth

I will answer your questions one by one.

Why do we need to have it calculated about a particular $λ_o$?

This directly follows from the definition of a derivative. I think you understand the meaning of derivative. Derivatives "at a point" are defined if the function is continuous in the neighbourhood of that point and also if the left and right derivatives exist and are equal. The point in this example happens to be $λ_o$. [One point to be noted : You can differentiate $nu=c/λ$ around the point $λ_o$. $nu=c/λ_o$ on differentiation yields zero, since the RHS is a constant].

Second question, this is true by definition, you can visualise this by drawing a tangent to a curve at a point and observing the $Delta$(a quantity) approaches d(a quantity). In fact d$nu$=lim $Delta nu$ as $Delta nu$ approaches zero.

Why do we disregard the negative sign because we chose to? How can this still justify the validity of our equation?

The negative sign appears because as frequency increases, the wavelength decreases and vice-versa. So, $Deltanu$ and $Deltalambda$ have opposite signs. We ignore the negative sign cautiously noting the relative increase or decrease of the wavelength.

Frequency ranges and wavelength ranges find useful applications. For instance, when an electron in an atom drops to lower energy states and emits a photon, there is an inherent uncertainty in the frequency of the emitted photon.