6.50. $\lim _{x \rightarrow 1} \frac{x^{4}-1}{\ln x}$.
Solution. Let's check the conditions for applying L'Hôpital's rule: 1) there is an indeterminate form

$$
\left.\frac{x^{4}-1}{\ln x}\right|_{x=1}=\frac{0}{0}
$$

2) the functions $x^{4}-1$ and $\ln x$ are differentiable in a neighborhood of the point $\left.x=1 ; 3)(\ln x)_{x=1}^{\prime} \neq 0 ; 4\right)$ the limit exists

$$
\lim _{x \rightarrow 1} \frac{\left(x^{4}-1\right)^{\prime}}{(\ln x)^{\prime}}=\lim _{x \rightarrow 1} \frac{4 x^{3}}{1 / x}=\lim _{x \rightarrow 1} 4 x^{4}=4
$$

Therefore, $\lim _{x \rightarrow 1} \frac{x^{4}-1}{\ln x}=\lim _{x \rightarrow 1} \frac{\left(x^{4}-1\right)^{\prime}}{(\ln x)^{\prime}}=4$.