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 
$$ f(x) = \frac{x^4-1}{x^3-1} . $$

Note that $x^4-1=(x-1)(x^3+x^2+x+1)$ and $x^3-1=(x-1)(x^2+x+1)$.
Thus
$$f(x)=\frac{(x-1)(x^3+x^2+x+1)}{(x-1)(x^2+x+1)}.$$
When $x\ne 1$, the $x-1$ terms cancel. Now we can safely let $x$ approach $1$.
