Let $[ t ]$ denote the greatest integer $\leq t$ and $\{ t \}$ denote the fractional part of $t$. Then integral value of $\alpha$ for which the left hand limit of the function $f ( x ) = [ 1 + x ] + \frac { \alpha ^ { 2 [ x ] + \{ x \} } + [ x ] - 1 } { 2 [ x ] + \{ x \} }$ at $x = 0$ is equal to $\alpha - \frac { 4 } { 3 }$ is $\_\_\_\_$