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Mathematics Question on limits and derivatives

Let [t] denote the greatest integer ≤ t and {t} denote the fractional part of t. The integral value of α for which the left-hand limit of the function f(x) = \left\lfloor 1 + x \right\rfloor + \frac{\alpha^{2\left\lfloor x \right\rfloor + \left\\{ x \right\\}} + \left\lfloor x \right\rfloor - 1}{2\left\lfloor x \right\rfloor + \left\\{ x \right\\}} at x=0 is equal to α−43\alpha -\frac{4}{3}, is

Answer

f(x) = \left\lfloor 1 + x \right\rfloor + \frac{\alpha^{2\left\lfloor x \right\rfloor + \left\\{ x \right\\}} + \left\lfloor x \right\rfloor - 1}{2\left\lfloor x \right\rfloor + \left\\{ x \right\\}}

lim⁡x→0−f(x)=α−43\lim_{{x \to 0^-}} f(x) = \alpha - \frac{4}{3}

⇒⇒ lim⁡x→0−[1+⌊x⌋+αx+⌊x⌋+⌊x⌋−1x+⌊x⌋]=α−43\lim_{{x \to 0^-}} \left[ 1 + \left\lfloor x \right\rfloor + \frac{\alpha^{x + \left\lfloor x \right\rfloor} + \left\lfloor x \right\rfloor - 1}{x + \left\lfloor x \right\rfloor} \right] = \alpha - \frac{4}{3}

⇒⇒ lim⁡h→0−[1−1+α−h−1−1−1−h−1]=α−43\lim_{{h \to 0^-}} \left[ 1 - 1 + \frac{\alpha^{-h - 1} - 1 - 1}{-h - 1} \right] = \alpha - \frac{4}{3}

∴∴ α−1−2−1=α−43\frac{\alpha^{-1} - 2}{-1} = \alpha - \frac{4}{3}
⇒⇒ 3α2–10α+3=03α^2 – 10α + 3 = 0
∴∴ α=3 or 13α = 3 \ or\ \frac{1}{3}
∵∵ α in integer, hence α=3α = 3