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 ______

We need to find the integral value of $$\alpha$$ for which the left-hand limit of $$f(x) = [1 + x] + \frac{\alpha^{2[x] + \{x\}} + [x] - 1}{2[x] + \{x\}}$$ at $$x = 0$$ equals $$\alpha - \frac{4}{3}$$.

For $$x$$ slightly less than 0, we have $$[x] = -1$$, $$\{x\} = x - [x] = x + 1$$, and $$[1 + x] = 0$$ since $$1 + x \to 1^-$$.

In this neighborhood, $$2[x] + \{x\} = 2(-1) + (x + 1) = x - 1 \to -1$$, $$\alpha^{2[x] + \{x\}} = \alpha^{x - 1} \to \alpha^{-1} = \frac{1}{\alpha}$$, and $$[x] - 1 = -1 - 1 = -2$$.

Substituting these into the definition of $$f$$ gives $$\lim_{x \to 0^-} f(x) = 0 + \frac{\frac{1}{\alpha} + (-2)}{-1} = 2 - \frac{1}{\alpha}$$.

Requiring this limit to equal $$\alpha - \frac{4}{3}$$ leads to the equation $$2 - \frac{1}{\alpha} = \alpha - \frac{4}{3}$$, which can be rewritten as $$\alpha + \frac{1}{\alpha} = \frac{10}{3}$$.

Multiplying through by $$\alpha$$ and rearranging yields the quadratic $$3\alpha^2 - 10\alpha + 3 = 0$$.

The solutions of this quadratic are $$\alpha = \frac{10 \pm \sqrt{100 - 36}}{6} = \frac{10 \pm 8}{6}$$, namely $$\alpha = 3$$ or $$\alpha = \frac{1}{3}$$.

Since we are looking for an integral value, the answer is $$\alpha = 3$$.

The correct answer is $$3$$.