Answer
$(a) \lim_{x\to1}\frac{x-1}{x^3-1} = \frac{1}{3},\\[6pt]
(b) \lim_{x\to1^+}\frac{x+1}{x^3-1} = +\infty,\\[6pt]
(c) \lim_{x\to1^-}\frac{x+1}{x^3-1} = -\infty.$
Work Step by Step
(a) $\lim_{x\to1}\frac{x-1}{x^3-1}$
Evaluating numerically:
\[
f(x)=\frac{x-1}{x^3-1}
\] \[
\begin{array}{c|cccccccccc}
x & 0 & 0.5 & 0.9 & 0.99 & 0.999 & 1.001 & 1.01 & 1.1 & 1.5 & 2 \\ \hline
f(x) & 1.000 & 0.571 & 0.369 & 0.337 & 0.334 & 0.333 & 0.330 & 0.302 & 0.211 & 0.143
\end{array}
\]
As \(x \to 1\), \(f(x)\) approaches approximately \(0.333\).
Algebraic (CAS) verification:
\[
x^3-1=(x-1)(x^2+x+1)
\]
\[
\frac{x-1}{x^3-1}=\frac{1}{x^2+x+1}, \quad x\ne1
\]
\[
\therefore \lim_{x\to1}\frac{x-1}{x^3-1}=\frac{1}{1^2+1+1}=\frac{1}{3}.
\]
\[
\boxed{\lim_{x\to1}\frac{x-1}{x^3-1}=\tfrac{1}{3}}
\]
(b) \(\displaystyle \lim_{x\to1^+}\frac{x+1}{x^3-1}\)}
Numerical evaluation:}
\[
\begin{array}{c|cccccc}
x & 1.0001 & 1.001 & 1.01 & 1.1 & 1.5 & 2 \\ \hline
f(x) & 6666.33 & 666.33 & 66.33 & 6.34 & 1.05 & 0.43
\end{array}
\]
As \(x\to1^+\), \(f(x)\to+\infty.\)
Reasoning:}
Since \(x^3-1=(x-1)(x^2+x+1)\), for \(x>1\), the denominator is a small positive number, and \(x+1>0\).
Hence, \(f(x)\) grows without bound:
\[
\boxed{\lim_{x\to1^+}\frac{x+1}{x^3-1}=+\infty.}
\]
(c) $lim_{x\to1^-}\frac{x+1}{x^3-1}$
Numerical evaluation:
\[
\begin{array}{c|cccccc}
x & 0 & 0.5 & 0.9 & 0.99 & 0.999 & 0.9999 \\ \hline
f(x) & -1.00 & -1.71 & -7.01 & -67.00 & -667.00 & -6667.00
\end{array}
\]
As \(x\to1^-\), \(f(x)\to-\infty.\)
(iii) Reasoning:
For \(x<1\), \(x-1<0\) and \(x^2+x+1>0\), so \(x^3-1<0\) while \(x+1>0\).
Thus, the quotient tends to a large negative number:
\[
\boxed{\lim_{x\to1^-}\frac{x+1}{x^3-1}=-\infty.}
\]
