Applying L'Hopital's Rule to a 0/0 limit

Substituting $x=1$ gives $\frac{1^3-1}{\ln(1)} = \frac{0}{0}$, which is an indeterminate form, so L'Hopital's Rule applies. Differentiating the numerator gives $\frac{d}{dx}[x^3-1] = 3x^2$, and differentiating the denominator gives $\frac{d}{dx}[\ln(x)] = \frac{1}{x}$.

By L'Hopital's Rule, the limit equals $\lim_{x\to 1} \frac{3x^2}{1/x} = \lim_{x\to 1} 3x^3.$ Substituting $x=1$ now gives $3(1)^3 = 3$, and since this is no longer indeterminate, this is the value of the original limit.

The answer is substitution. After applying L'Hopital's Rule and differentiating the numerator and denominator separately, the resulting expression $3x^3$ is no longer indeterminate at $x = 1$. So the final step is simply substituting $x = 1$ into this new expression to get the value $3$. No further differentiation or algebraic manipulation is needed.