Explaining the optial behaviour of DVD reader lens Here is the lens removed from a DVD reader/writer's head

I do not have the dimension of the lens, neither could I find it online. Correct me if I am wrong, but I guess it is a plano-convex lens.
I placed this lens on top of a cellphone screen(with the flat side coinciding with the screen) and here is what I saw - a highly magnified image of the RGB pixels of the screen.
Here is a collage of screenshots from a video -

Optically, the object is placed very close to the lens. Then using the lens formula
$$
{1 \over v} - {1 \over u} = {1 \over f}
$$
we get $v = u$ for $u =0$. And by the magnification equation,
$$
M = -{v \over u}
$$
we get magnification as $1$ for $\lim {u \to 0}$. But we see a very large magnification. How to explain this optical behaviour?
 A: First, a few links will help. Here is a quick overview of what lens design is. Here is an explanation of how lenses work. The second link has several useful links. Start with magnification. You can see that a large magnification requires light rays to be deflected by a large angle. 
You are getting a large magnification because is this is a strong lens (short focal length). You can tell because the lens is strongly curved. The spherical side has a small radius. Large diameter lenses cannot be this strong. 
There are various ways to calculate the focal length of a lens. The simplest is the thin lens approximation. It is for cases where the thickness of the lens is small compared to other lengths (weaker lenses). You cannot use it for this lens. 
For a thick lens, you need to take the location of each surface into account when calculating focal length. The thick lens formula is a better approximation. 
For a thick lens, you can still use $1/v - 1/u = 1/f$. But you need to measure from the right locations, as shown here. 
So u is not 0. You would have to know the lens parameters to calculate it. 
