If object size increases, shouldn't magnification decrease by formula According to the magnification formula, magnification is the image size by object size
According to this if object comes near and its size increases shouldn't magnification decrease? 
 A: The answer by @QuIKmAtHs is not correct.
The lensmaker's formula: $$1/d_1 + 1/d_2 = 1/f$$ is a good starting point.  The magnification is, indeed, $$ M = d_2/d_1;$$ but $d_2$ does not decrease when $d_1$ decreases; it typically increases when $d_1$ decreases.  When $d_1 = f$, the magnification is essentially infinite.  
When you see a lens labeled "5x", the 5x refers to angular magnification, not magnification of the size of the image relative to the object.
A: Magnification is an inherent property of the lens, independent of the height and distance of the original object.  Let’s say we are moving a toy car towards the lens, then the distance between object and the lens decreases, but the distance between the image decreases too.  The formula is $M=\frac{d_i}{d_o}$.
On the other hand, if your object is larger, then the image becomes larger (higher).    $M=\frac{H_i}{H_o}$ is the formula for magnification.  Hence, these two factors are independent of the magnification:  the magnification of a lens is constant.
A: Magnification of the focused object can be shown to be:
$m = \frac{f}{s-f}$,
where $f$ is the focal distance and $s$ distance from the object to the lens. So if you keep your subject in focus as it comes closer, magnification will indeed increase, not decrease.
In case you only focus the lens once and then move the object around without changing that focus, there is a bit different formula for a defocused object:
$m_d = \frac{s m}{s_d}$,
where $m$ is the magnification at the focused point, $s$ the distance of that focused spot, and $m_d$ and $s_d$ magnification and distance at the defocused spot. As you can see, the relationship still stays the same: magnification increases as the object gets closer.
