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> What justifies using L'Hopitals Rule for $\lim\limits\_{x\to1}\frac{x^3-x}{\ln(x-1)}$?
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I figured that the limit can be computed like that:$$\lim\limits\_{x\to1 }\frac{x^3-x}{\ln(x-1)}=\lim\limits\_{x\to1 }\frac{3x^2-1}{\frac{1}{x-1}}=\lim\limits\_{x\to 1}( 3x^2-1)(x-1)=0$$ But why can we use L'Hopital? We neither have $0/0$ or $\infty/\infty$