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