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Step-by-Step Solution
Step 1: Understand the function and the limit
The function given is
f(x) = [\,1 + x\,] + \dfrac{\alpha^{2[x] + \{x\}} + [x] - 1}{2[x] + \{x\}} ,
where [\,t\,] denotes the greatest integer less than or equal to t , and \{\,t\,\} = t - [\,t\,] denotes the fractional part of t .
We need to find the integral value of \alpha such that
\displaystyle \lim_{x \to 0^-} f(x) = \alpha - \tfrac{4}{3} .
Step 2: Determine behavior of [x] and {x} as x → 0⁻
As x \to 0 from the left (negative side), we have:
[\,x\,] = -1 because any negative x close to 0 will have integer part -1.
\{\,x\,\} = x - [\,x\,] = x - (-1) = x + 1.
Thus, for x approaching 0 from the left:
\[
2[x] + \{x\} \;=\; 2(-1) + (1 + x) \;=\; -2 + 1 + x \;=\; -1 + x.
\]
Since x \to 0^- , this expression approaches -1 (slightly less than -1 as x is negative and close to 0).
Step 3: Rewrite the limit using these values
Substituting [\,x\,] = -1 and \{\,x\,\} = x + 1 into the expression, we get:
\[
f(x) \;=\; [\,1 + x\,] + \dfrac{\alpha^{\,2[x] + \{x\}} + [x] - 1}{2[x] + \{x\}}.
\]
For x \to 0^- , [\,1 + x\,] = 1 + [\,x\,] because 0^-\!< 1 + x < 1 ; more exactly, [\,1 + x\,] can be 0 or 1 depending on how close x is to -1. However, commonly for x just a bit less than 0, 1 + x is slightly less than 1, so [\,1 + x\,] = 0 .
But the solution steps given simplify it directly by focusing on the limit and combining terms carefully. One standard approach (as shown in the reference solution) is to set up a small variable h such that x = -h with h \to 0^+ , then track the expression carefully.
Step 4: Simplify and take the limit
In the provided reference simplification, the limit transforms into
\[
\lim_{x \to 0^-}
\left(
[\,1 + x\,] + \dfrac{\alpha^{\,2[x] + \{x\}} + [x] - 1}{2[x] + \{x\}}
\right)
=
\lim_{x \to 0^-}
\left(
1 + [x] + \dfrac{\alpha^{\,x + [x]} + [x] - 1}{x + [x]}
\right),
\]
where we note 2[x] + \{x\} = x + [x] because 2[x] + \{x\} = 2[x] + x - [x] = [x] + x .
Substituting [\,x\,] = -1 into x + [x] gives x - 1 . As x \to 0^- , x - 1 \to -1 . Eventually, the reference work leads to:
\[
\lim_{x \to 0^-}
\dfrac{\alpha^{-1} - 2}{-1}
= \alpha - \dfrac{4}{3}.
\]
Carefully handling exponents and expansions shows that we reach a quadratic equation in \alpha .
Step 5: Solve the resulting equation for α
From the reference derivation, the condition becomes:
\[
\dfrac{\alpha^{-1} - 2}{-1} = \alpha - \dfrac{4}{3}.
\]
Manipulating further yields a quadratic:
\[
3 \alpha^2 - 10 \alpha + 3 = 0.
\]
Solving this quadratic equation gives two possible solutions for \alpha :
\[
\alpha = 3 \quad \text{or} \quad \alpha = \tfrac{1}{3}.
\]
However, the problem asks specifically for the integral value of \alpha . Therefore,
\alpha = 3 is the required integer solution.
Final Answer
The integral value of \alpha that satisfies the condition is 3.
