\( \lim \limits_{x \to 1} \dfrac{2\cdot \ln(x)}{e^x-1}\)

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$$ \lim \limits_{x \to 1} \frac{2\cdot \ln(x)}{e^x-1}$$ $$ = \frac{2\ln(1)}{e^1-1} $$ $$ = \frac{2(0)}{e-1} $$ $$ = \frac{0}{e-1} $$ $$ = \boxed{0} $$