A) \[\log 2\]
B) \[\log 4\]
C) \[\log \sqrt{2}\]
D) None of these
Correct Answer: B
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{2}^{x}}-1}{{{(1+x)}^{1/2}}-1}=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{2}^{x}}\log 2}{\tfrac{1}{2}\,{{(1+x)}^{-1/2}}}\] \[\left\{ \because \,\,\,\underset{x\to a}{\mathop{\lim }}\,\,\frac{f(x)}{g(x)}=\underset{x\to a}{\mathop{\lim }}\,\,\frac{{f}'(x)}{{g}'(x)} \right\}\] \[=2\,\log 2=\log 4.\]
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