Key Concepts and Formulas

• Greatest Integer Function $[x]$: The greatest integer less than or equal to $x$. For $x \to 0^-$, if $x$ is a small negative number (e.g., $-0.001$), then $[x] = -1$.
• Fractional Part Function $\{x\}$: Defined as $\{x\} = x - [x]$. For $x \to 0^-$, $\{x\} = x - [x] = x - (-1) = x+1$. As $x \to 0^-$, $\{x\} \to 1$.
• Left-Hand Limit: The limit of a function as the variable approaches a certain value from the left side. Notation: $\mathop {\lim }\limits_{x \to a^-} f(x)$.
• Limits of Exponential Functions: For a base $a > 0$, $\mathop {\lim }\limits_{y \to c} a^y = a^c$.

Step-by-Step Solution

Step 1: Analyze the function near $x = 0^-$ We are given the function f(x) = [1 + x] + {{{{\alpha }^{2[x] + \{x\}}}} + [x] - 1} \over {2[x] + \{ x\} }}. We need to find the left-hand limit of $f(x)$ as $x \to 0^-$. As $x \to 0^-$, $x$ takes small negative values. Therefore, $[x] = -1$. And $\{x\} = x - [x] = x - (-1) = x+1$. As $x \to 0^-$, $\{x\} \to 1$. Also, $1+x \to 1+0^- = 1^-$, so $[1+x] = 0$.

Step 2: Substitute the values of $[x]$ and $\{x\}$ into the function for the limit calculation. For $x \to 0^-$, we have $[x] = -1$ and $\{x\} = x+1$. The expression for $f(x)$ becomes: f(x) = [1+x] + {{{{\alpha }^{2(-1) + (x+1)}}}} + (-1) - 1} \over {2(-1) + (x+1)}} f(x) = 0 + {{{{\alpha }^{-1 + x}}}} + x - 2} \over {-2 + x + 1}} f(x) = {{{{\alpha }^{x-1}}}} + x - 2} \over {x - 1}}

Step 3: Evaluate the left-hand limit of the simplified function. We need to calculate $\mathop {\lim }\limits_{x \to {0^ - }} f(x)$. \mathop {\lim }\limits_{x \to {0^ - }} f(x) = \mathop {\lim }\limits_{x \to {0^ - }} {{{{\alpha }^{x-1}}}} + x - 2} \over {x - 1}} Now, substitute $x=0$ into the expression since the denominator will not be zero and the numerator will be well-defined. \mathop {\lim }\limits_{x \to {0^ - }} f(x) = {{{{\alpha }^{0-1}}}} + 0 - 2} \over {0 - 1}}} \mathop {\lim }\limits_{x \to {0^ - }} f(x) = {{{{\alpha }^{-1}}}} - 2} \over {-1}} $\mathop {\lim }\limits_{x \to {0^ - }} f(x) = 2 - \alpha^{-1}$

Step 4: Set the left-hand limit equal to the given value. We are given that the left-hand limit is equal to $\alpha - {4 \over 3}$. So, $2 - \alpha^{-1} = \alpha - {4 \over 3}$.

Step 5: Solve the equation for $\alpha$. $2 - {1 \over \alpha} = \alpha - {4 \over 3}$ Multiply the entire equation by $3\alpha$ to eliminate denominators: $3\alpha(2) - 3\alpha({1 \over \alpha}) = 3\alpha(\alpha) - 3\alpha({4 \over 3})$ $6\alpha - 3 = 3\alpha^2 - 4\alpha$ Rearrange the terms to form a quadratic equation: $3\alpha^2 - 4\alpha - 6\alpha + 3 = 0$ $3\alpha^2 - 10\alpha + 3 = 0$

Step 6: Find the roots of the quadratic equation. We can factor the quadratic equation $3\alpha^2 - 10\alpha + 3 = 0$. We look for two numbers that multiply to $3 \times 3 = 9$ and add up to $-10$. These numbers are $-9$ and $-1$. So, $3\alpha^2 - 9\alpha - \alpha + 3 = 0$ $3\alpha(\alpha - 3) - 1(\alpha - 3) = 0$ $(\alpha - 3)(3\alpha - 1) = 0$ The possible values for $\alpha$ are $\alpha = 3$ or $\alpha = {1 \over 3}$.

Step 7: Select the integral value of $\alpha$. The question asks for the integral value of $\alpha$. Between $\alpha = 3$ and $\alpha = {1 \over 3}$, only $\alpha = 3$ is an integer.

Common Mistakes & Tips

• Handling $[x]$ and $\{x\}$ for $x \to 0^-$: Be very careful with the values of $[x]$ and $\{x\}$ as $x$ approaches 0 from the left. $[x]$ becomes $-1$, and $\{x\}$ approaches $1$.
• Algebraic Errors in Solving the Equation: Double-check all algebraic manipulations, especially when clearing denominators and solving the quadratic equation.
• Ignoring the Integral Constraint: The problem specifically asks for an integral value of $\alpha$. Ensure that the final answer satisfies this condition.

Summary The problem requires calculating the left-hand limit of a function involving the greatest integer and fractional part functions at $x=0$. By carefully analyzing the behavior of $[x]$ and $\{x\}$ as $x \to 0^-$, we simplified the function and evaluated its limit. Equating this limit to the given expression $\alpha - {4 \over 3}$ led to a quadratic equation in $\alpha$. Solving this equation gave two possible values for $\alpha$, and selecting the integer value provided the final answer.

The final answer is $\boxed{3}$.