question_answer

Examine the continuity of \[f(x)=\left\{ \begin{matrix}    \frac{\log x-\log 2}{x-2} & x>2  \\    \frac{1}{2}, & x=2\,\,\text{at}\,\,x=2.  \\    2\left( \frac{x-2}{{{x}^{2}}-4} \right), & x<2  \\ \end{matrix} \right.\]            

Answer:

\[A+x=2,\] \[LHL=\underset{h\to {{2}^{-}}}{\mathop{\lim }}\,f(x)\] \[=\underset{h\to 0}{\mathop{\lim }}\,(2-h)=\underset{h\to 0}{\mathop{\lim }}\,\frac{2\{(2-h)-2\}}{\{{{(2-h)}^{2}}-4\}}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{-\,2h}{4+{{h}^{2}}-4h-4}=\underset{h\to 0}{\mathop{\lim }}\,\frac{-\,2}{h-4}=\frac{1}{2}\] and \[RHL=\underset{x\to {{2}^{+}}}{\mathop{\lim }}\,f(x)\]             \[=\underset{h\to 0}{\mathop{\lim }}\,(2+h)=\underset{h\to 0}{\mathop{\lim }}\,\frac{\log (2+h)-\log 2}{2+h-2}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{log\left( 1+\frac{h}{2} \right)}{h}\]         \[\left[ \because \,\,\log m-\log n=\log \frac{m}{n} \right]\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{log\left( 1+\frac{h}{2} \right)}{2\left( \frac{h}{2} \right)}=\frac{1}{2}\] \[\left[ \because \,\,\,\,\underset{h\to 0}{\mathop{\lim }}\,\frac{\log (1+x)}{x}=1 \right]\] Also, \[f(2)=\frac{1}{2}\]            Thus, LHL = RHL = f(2) \[\therefore \] f(x) is continuous at x = 2.