A) \[\underset{x\to \frac{{{\pi }^{-}}}{2}}{\mathop{\lim }}\,\,\,f(x)\,=3\]
B) \[\underset{x\to \frac{{{\pi }^{+}}}{2}}{\mathop{\lim }}\,\,\,f(x)\,=0\]
C) \[\underset{x\to \frac{{{\pi }^{+}}}{2}}{\mathop{\lim }}\,\,\,f(x)\,=3\]
D) \[\underset{x\to \frac{\pi }{2}}{\mathop{\lim }}\,\,\,f(x)\] exists
Correct Answer: C
Solution :
\[f(x)=\frac{3}{1+{{3}^{\tan x}}}\] LHL at \[\left( x=\frac{\pi }{2} \right),\] \[f\left( \frac{\pi }{2}-0 \right)=\underset{h\to 0}{\mathop{\lim }}\,\,f\left( \frac{\pi }{2}-h \right)=\underset{h\to 0}{\mathop{\lim }}\,\frac{3}{1+{{3}^{\tan \left( \frac{\pi }{2}-h \right)}}}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{3}{1+{{3}^{\cot \,h}}}=\frac{3}{1+{{3}^{\cot \,0}}}\] \[\Rightarrow \] \[\frac{3}{1+{{(3)}^{\infty }}}=\frac{3}{1+\infty }=0\] RHL at \[\left( x=\frac{\pi }{2} \right),\] \[f\left( \frac{\pi }{2}+0 \right)=\underset{h\to 0}{\mathop{\lim }}\,\,f\left( \frac{\pi }{2}+h \right)=\underset{h\to 0}{\mathop{\lim }}\,\,\frac{3}{1+{{3}^{\tan \left( \frac{\pi }{2}+h \right)}}}\] \[=\underset{h\to 0}{\mathop{\lim }}\,\frac{3}{1+{{3}^{-cot\,h}}}=\frac{3}{1+{{3}^{-\cot \,0}}}\] \[=\frac{3}{1+{{3}^{-\infty }}}=\frac{3}{1+0}=3\] Hence, \[\underset{x\to \frac{{{\pi }^{+}}}{2}}{\mathop{\lim }}\,\,\,f(x)=3\]
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