Numerically evaluating the limit of $\frac{x^4-1}{x^3-1}$ as $x\rightarrow 1$ What is the limit as $x \to 1$ of the function 
$$ f(x) = \frac{x^4-1}{x^3-1} . $$

Applying L'Hôpital's rule makes it quite straightforward.
$$ \lim_{x \rightarrow 1}\; \frac{x^4-1}{x^3-1} = \lim_{x \rightarrow 1} \;\frac{4x^3}{3x^2} =  \lim_{x \rightarrow 1} \;\frac{4}{3}x = \frac{4}{3}$$
