Chemical Equilibrium - Le Chatelier's principle
For the case you're describing, having left-over product species will likely not cause the reverse reaction to increase.

When you define a reaction equation like: $\ce{aA + bB &lt;=&gt; cC + dD}$, you are explicitly defining A and B as 'reactants' and C and D as 'products'. In the scenario you've described, just because you have excess reactants because you have a limiting concentration of either A or B, doesn't mean the concentration of products will increase. This is because we can think of concentration of products as $[Products](t) = [C](t) + [D](t)$ and not $[Products](t) = [C](t) + [D](t) + [A](t) + [B](t)$.

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A nice definition of Le Chatlier's principle is the following: "A change in one of the variables that describe a system at equilibrium produces a shift in the position of the equilibrium that counteracts the effect of this change."

If one is to get a little more technical with this, equilibrium is a function of many factors, primarily Temperature (T), Pressure (P), and mixture species (n). This is typically described by thermodynamics by the state property, Gibbs Free Energy, or $\Delta G$.

For the purposes of an example to quantify Le Chatlier's principle, let's look at the differential form of Gibbs Free Energy: $dG = -SdT +VdP + \sum^{n}_{i=1}\mu_idn_i$. If we consider a case where temperature isn't changing (isothermal), and where pressure isn't changing (isobaric), then the Gibbs Free Energy equation is affected by changes in system composition: $dG = \sum^{n}_{i=1}\mu_idn_i$. For an equilibrium case, which means that $\Delta G = 0$, it can be shown that if you expand this expression out for a reaction mechanism (like the one mentioned above), you end up with an expression like: $\frac{\Delta G^o_{T_o}}{RT_o} = -ln(K_{eq})$. This particular equation is defined for standard conditions (0 degC, 1 atm).

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We're almost to an answer, let's look a little closer at $K_{eq}$. Let's use a gas phase system $\ce{aA + bB &lt;=&gt; cC + dD}$ as an example. If you look up the derivation of the $\frac{\Delta G^o_{T_o}}{RT_o} = -ln(K_{eq})$ in any standard thermodynamics textbook, you will see that $K_{eq}$ is actually equal to the ratio of the mol fractions of each species multiplied by the system pressure: $K_{eq} = \frac{y_C^c y_D^d}{y_A^a y_B^b} P^{c + d - a - b}$. If one assumes that the pressure is fixed and that $a = b = c = d = 1$, then this expression would reduce down to: $K_{eq} = \frac{y_C y_D}{y_A y_B}$.

You can further add detail to this expression by defining the mol fraction in terms of the extent of reaction $\xi$ for a reaction system. 
Where, $n_i = n_{io} + v_in_{io}\xi$--which essentially says that the mols of a given species is equal to inital mols +/- the extent to which that chemical is generated or reacted away (based on the specified chemical of reference $n_{io}$). If you plug this into the expression for $K_{eq}$ you can see that now the equilibrium conversion of a reaction system is dependent on the amount of mols of each species. For the system described above, the total mols $N_{total}$ will cancel out of the expression leaving:

$K_{eq} = \frac{(n_{Ao} + v_An_{io})(n_{Bo} + v_Bn_{io})}{(n_{Co} + v_CN_{io})(n_{Do} + v_Dn_{io})} = exp(-\frac{\Delta G^o_{T_o}}{RT_o})$

For a given temperature and pressure, the Gibbs Free Energy ($\Delta G$) will be a fixed value. Therefore the conditions in the equation (composition of the mixture of reaction system) must balance in such a way that the equilibrium conversion $\xi$ will satisfy the condition. You can see here that as one either varies products (C,D) or reactants (A,B) that the solution for $\xi$ will change. You can generally show that an excess in either products or reactants will shift the system to maximum equilibrium conversion $\xi$. And as such, if one suddenly added or removed products or reactants you would cause a shift in the equilibrium conversion $\xi$.