Answer
$$\lim _{x \rightarrow 1} \frac{x^{4}-1}{x^{3}-1} = \frac{4}{3}$$
Work Step by Step
Given $$\lim _{x \rightarrow 1} \frac{x^{4}-1}{x^{3}-1}$$
using the Limit Rules and replacement, leads to the indeterminate form $$\lim _{x \rightarrow 1} \frac{x^{4}-1}{x^{3}-1}=\frac{0}{0}$$
Applying L'Hôpital's Rule
\begin{align*}
\lim _{x \rightarrow 1} \frac{x^{4}-1}{x^{3}-1}&=\lim _{x \rightarrow 1} \frac{4x^{3}}{3x^{2}}\\
&=\lim _{x \rightarrow 1} \frac{4}{3}x \\
&=\lim _{t \rightarrow \infty} \frac{4}{3}\\
&= \frac{4}{3}
\end{align*}
