What is the largest gap between rank and approximate rank?
First I'll give some background and define approximate rank. A good reference is the recent survey by Lee and Schraibman Lower Bounds on Communication Complexity.

Definition: Let $A$ be a sign matrix. The approximate rank of $A$ with approximation factor $\alpha$, denoted $rank^\alpha(A)$, is
$rank^\alpha(A)=\min_{B:1\leq A[i,j]\cdot B[i,j] \leq \alpha} rank(B)$.
When $\alpha\to \infty$, define
$rank^\alpha(A)=\min_{B:1\leq A[i,j]\cdot B[i,j]} rank(B)$.

A result by Krause says that $R_\epsilon^{pri}(A)\geq \log rank^\alpha(A)$ where $\alpha=1/(1-2\epsilon)$ and $R_\epsilon^{pri}$ is the bounded-error private-coin communication complexity of $A$ with error upper-bounded by $\epsilon$.
The above was for background. Now to answer the question, Paturi and Simon showed that $rank^\infty(A)$ completely characterizes the unbounded-error communication complexity of $A$. They also showed that this agrees with the minimum dimension of an arrangement realizing the boolean function whose communication matrix is $A$. The unbounded-error communication complexity of the equality function is $O(1)$. Keep that in mind.
The communication matrix for equality is just the identity, i.e., a boolean matrix with $2^n$ rows and $2^n$ columns with all ones in the diagonal. Let's denote this by $I_{2^n}$. Alon showed that $rank^2(I_{2^n})=\Omega(n)$ which is tight up to a logarithmic factor (with the theorem by Krause we obtain $R_\epsilon^{pri}(EQ)=\Omega(\log n)$).
The identity matrix has full rank, i.e., $2^n$. Thus, we have exponentially large separations for $\alpha=2$ and $\alpha\to\infty$.