Answer
$$\lim _{x \rightarrow 1} \frac{x^{2}-5 x+4}{x^{3}-1}=-1$$
Work Step by Step
Given $$\lim _{x \rightarrow 1} \frac{x^{2}-5 x+4}{x^{3}-1}$$
let $$ f(x) = \frac{x^{2}-5 x+4}{x^{3}-1}$$
Since, we have
$$ f( 1)= \frac{1-5+4}{1-1}=\frac{0}{0}$$
So, transform algebraically and cancel, we get
\begin{aligned}
L&=\lim _{x \rightarrow 1} \frac{x^{2}-5 x+4}{x^{3}-1}\\
&=\lim _{x \rightarrow 1} \frac{(x-1)(x-4)}{(x-1)\left(x^{2}+x+1\right)}\\
&=\lim _{x \rightarrow 1} \frac{x-4}{x^{2}+x+1}\\
&=\frac{1-4 }{1+1+1} \\
&=-\frac{3 }{3} \\
&=-1
\end{aligned}
