Answer
$$1$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to {1^ + }} \left( {\frac{2}{{\ln x}} - \frac{2}{{x - 1}}} \right) \cr
& {\text{Evaluating the limit}} \cr
& \mathop {\lim }\limits_{x \to {1^ + }} \left( {\frac{2}{{\ln x}} - \frac{2}{{x - 1}}} \right) = \frac{2}{{\ln \left( {{1^ + }} \right)}} - \frac{2}{{\left( {{1^ + }} \right) - 1}} = \infty - \infty \cr
& {\text{Reduce the rational functions}} \cr
& = \mathop {\lim }\limits_{x \to {1^ + }} \left( {\frac{{2\left( {x - 1} \right) - 2\ln x}}{{\ln x\left( {x - 1} \right)}}} \right) \cr
& {\text{Using L'Hopital's rule}} \cr
& = \mathop {\lim }\limits_{x \to {1^ + }} \left( {\frac{{\frac{d}{{dx}}\left[ {2\left( {x - 1} \right) - 2\ln x} \right]}}{{\frac{d}{{dx}}\left[ {\ln x\left( {x - 1} \right)} \right]}}} \right) \cr
& = \mathop {\lim }\limits_{x \to {1^ + }} \left( {\frac{{2 - \frac{2}{x}}}{{\ln x - \frac{{x - 1}}{x}}}} \right) \cr
& {\text{Evaluating the limit}} \cr
& = \mathop {\lim }\limits_{x \to {1^ + }} \left( {\frac{{2 - \frac{2}{1}}}{{\ln 1 - \frac{{1 - 1}}{1}}}} \right) = \frac{0}{0} \cr
& {\text{Using L'Hopital's rule}} \cr
& = \mathop {\lim }\limits_{x \to {1^ + }} \left( {\frac{{\frac{d}{{dx}}\left[ {2 - \frac{2}{x}} \right]}}{{\frac{d}{{dx}}\left[ {\ln x + 1 - \frac{1}{x}} \right]}}} \right) \cr
& = \mathop {\lim }\limits_{x \to {1^ + }} \left( {\frac{{\frac{2}{{{x^2}}}}}{{\frac{1}{x} + \frac{1}{{{x^2}}}}}} \right) \cr
& {\text{Evaluating the limit}} \cr
& = \frac{{\frac{2}{{{{\left( 1 \right)}^2}}}}}{{\frac{1}{{\left( 1 \right)}} + \frac{1}{{{{\left( 1 \right)}^2}}}}} = \frac{2}{2} = 1 \cr
& {\text{Therefore,}} \cr
& \mathop {\lim }\limits_{x \to {1^ + }} \left( {\frac{2}{{\ln x}} - \frac{2}{{x - 1}}} \right) = 1 \cr} $$
