Answer
$$ 3-a$$
Work Step by Step
By substituting, we get the intermediate form $\frac{0}{0}$. So by factorizing the numerator, we have
$$
\lim _{x \rightarrow 1} \frac{ x^3-ax^2+ax-1} {x-1 }=\lim _{x \rightarrow 1} \frac{( x^3- 1)-ax(x-1)} {x-1 }\\
=\lim _{x \rightarrow 1} \frac{( x- 1)(x^2+x+1)-ax(x-1) }{x-1 }=\lim _{x \rightarrow 1} x^2+x -ax+1\\
=1+1-a+1=3-a.
$$
