Answer
$ a.\quad +\infty$
$ b.\quad -\infty$
$ c.\quad -\infty$
$ d.\quad -\infty$
Work Step by Step
$ a.\qquad x\rightarrow 0^{+}\Rightarrow$
$\displaystyle \left[\begin{array}{l}
x^{1/3}\rightarrow 0^{+}\\
\\
\frac{1}{x^{1/3}}\rightarrow+\infty
\end{array}\right],\ \ \left[\begin{array}{l}
(x-1)^{4/3}=[(x-1)^{1/3}]^{4}\rightarrow+1\\
\\
\frac{1}{(x-1)^{4/3}}\rightarrow+1
\end{array}\right]\ \ \Rightarrow(\frac{1}{x^{1/3}}-\frac{1}{(x-1)^{4/3}})\rightarrow+\infty$
$ b.\qquad x\rightarrow 0^{-}\Rightarrow$
$\displaystyle \left[\begin{array}{l}
x^{1/3}\rightarrow 0^{-}\\
\\
\frac{1}{x^{1/3}}\rightarrow-\infty
\end{array}\right],\ \ \left[\begin{array}{l}
(x-1)^{4/3}=[(x-1)^{1/3}]^{4}\rightarrow+1\\
\\
\frac{1}{(x-1)^{4/3}}\rightarrow+1
\end{array}\right]\ \ \Rightarrow(\frac{1}{x^{1/3}}-\frac{1}{(x-1)^{4/3}})\rightarrow-\infty$
$ c.\qquad x\rightarrow 1^{+}\Rightarrow$
$\displaystyle \left[\begin{array}{l}
x^{1/3}\rightarrow+1\\
\\
\frac{1}{x^{1/3}}\rightarrow+1
\end{array}\right],\ \ \left[\begin{array}{l}
(x-1)^{4/3}=[(x-1)^{1/3}]^{4}\rightarrow 0^{+}\\
\\
\frac{1}{(x-1)^{4/3}}\rightarrow+\infty
\end{array}\right]\ \ \Rightarrow(\frac{1}{x^{1/3}}-\frac{1}{(x-1)^{4/3}})\rightarrow-\infty$
$ d.\qquad x\rightarrow 1^{-}\Rightarrow$
$\displaystyle \left[\begin{array}{l}
x^{1/3}\rightarrow+1\\
\\
\frac{1}{x^{1/3}}\rightarrow+1
\end{array}\right],\ \ \left[\begin{array}{l}
(x-1)^{4/3}=[(x-1)^{1/3}]^{4}\rightarrow 0^{+}\\
\\
\frac{1}{(x-1)^{4/3}}\rightarrow+\infty
\end{array}\right]\ \ \Rightarrow(\frac{1}{x^{1/3}}-\frac{1}{(x-1)^{4/3}})\rightarrow-\infty$