Henry decides one morning to do a workout, and he walks $\tfrac{3}{4}$ of the way from his home to his gym. The gym is $2$ kilometers away from Henry's home. At that point, he changes his mind and walks $\tfrac{3}{4}$ of the way from where he is back toward home. When he reaches that point, he changes his mind again and walks $\tfrac{3}{4}$ of the distance from there back toward the gym. If Henry keeps changing his mind when he has walked $\tfrac{3}{4}$ of the distance toward either the gym or home from the point where he last changed his mind, he will get very close to walking back and forth between a point $A$ kilometers from home and a point $B$ kilometers from home. What is $|A-B|$?
$\textbf{(A) } \frac{2}{3} \qquad \textbf{(B) } 1 \qquad \textbf{(C) } 1 \frac{1}{5} \qquad \textbf{(D) } 1 \frac{1}{4} \qquad \textbf{(E) } 1 \frac{1}{2}$
Let the two points that Henry walks in between be $P$ and $Q$, with $P$ being closer to home. As given in the problem statement, the distances of the points $P$ and $Q$ from his home are $A$ and $B$ respectively. By symmetry, the distance of point $Q$ from the gym is the same as the distance from home to point $P$. 
Thus, $A = 2 - B$. 
In addition, when he walks from point $Q$ to home, he walks $\frac{3}{4}$ of the distance, ending at point $P$. Therefore, we know that $B - A = \frac{3}{4}B$. 
By substituting, we get $B - (2-B) = \frac{3}{4}\cdot B$ and we solve to get $B=\dfrac{8}{5}$, so $A=2-\dfrac{8}{5}=\dfrac{2}{5}$. 
$|A-B|=\left|\dfrac{2}{5}-\dfrac{8}{5} \right|=\frac{6}{5}=\boxed{\textbf{(C) } 1 \frac{1}{5}}$.