Compute, $\lim_{x \to 2 }\, \frac{{{\log }_{2}}x-{{\log }_{x}}2}{\ln x-\ln 2}$

We know that , ${{\log}_{a}}b =\frac{\ln b}{\ln a}$ where $a,b \gt 0\,\And \,a\ne 1$

So $\frac{{{\log}_{2}}x-{{\log }_{x}}2}{\ln x-\ln 2}$

$=\frac{\frac{\ln x}{\ln 2}-\frac{\ln 2}{\ln x}}{\ln x-\ln 2}$

$=\frac{\frac{\ln x\ln x-\ln 2\ln 2}{\ln 2\ln x}}{\ln x-\ln2}$

$=\frac{{{\ln }^{2}}x-{{\ln }^{2}}2}{\left( \ln x-\ln 2 \right)\ln 2\ln x}$

$=\frac{\left(\ln x-\ln 2 \right)\left( \ln x+\ln 2 \right)}{\left( \ln x-\ln 2 \right)\ln 2\ln x}$

So $\underset{x\to2}{\mathop{\lim }}\,\frac{{{\log }_{2}}x-{{\log }_{x}}2}{\ln x-\ln 2}$

$=\underset{x\to 2}{\mathop{\lim }}\,\frac{\ln x+\ln 2}{\ln 2\ln x}$

$=\frac{\ln2+\ln 2}{\ln 2\ln 2}$

$=\frac{\ln \left( 4 \right)}{{{\left( \ln 2\right)}^{2}}}$

$=\frac{2\ln 2}{{{\left( \ln 2 \right)}^{2}}}$

$=\frac{2}{\ln 2}$