A) \[\frac{1}{2}\log 2\]
B) \[\log 4\]
C) \[\log 8\]
D) \[\log 2\]
E) none of these
Correct Answer: B
Solution :
\[\because \]\[f(x)=\left\{ \begin{matrix} \frac{{{2}^{x}}-1}{\sqrt{1+x-1}}, & x\ne 0 \\ k, & x=0 \\ \end{matrix} \right.\] \[\therefore \,\underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{2}^{x}}-1}{\sqrt{1+x}-1}\,=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{2}^{x}}\,{{\log }_{e}}2}{\frac{1}{2\sqrt{1+x}}}\] (By LHospitals rule) \[=2{{\log }_{e}}2={{\log }_{e}}4\] \[\because \]\[f(x)\]is continuous at\[k=0\] \[\therefore \] \[\underset{x\to 0}{\mathop{\lim }}\,f(x)=f(0)\] \[\Rightarrow \] \[{{\log }_{e}}4=k\]
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