$\color{red}{f(x) = L\cdot x}$ torna $f$ definida nos reais e respeita o limite $\underset{x\space \rightarrow \space 0 }{\lim }\space \displaystyle{\frac{f(x)}{x}} = L$ . Assim:$$\underset{x\space \rightarrow \space 1 }{\lim }\space \displaystyle{\frac{f(x^2-1)}{x-1}} = \underset{x\space \rightarrow \space 1 }{\lim }\space \displaystyle{\frac{L\cdot (x-1)(x+1)}{x-1}} = \underset{x\space \rightarrow \space 1 }{\lim }\space \displaystyle{L\cdot (x+1)}$$$$\boxed{\underset{x\space \rightarrow \space 1 }{\lim }\space \displaystyle{L\cdot (x+1)} \space =\space 2L}$$$$\text{Alternativa } \mathbb{(E)}$$