A) 0
B) \[\log 4\]
C) \[\log 2\]
D) None of these
Correct Answer: B
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{x.({{2}^{x}}-1)}{1-\cos x}=\underset{x\to 0}{\mathop{\lim }}\,\frac{{{2}^{x}}-1}{x}.\frac{{{x}^{2}}}{1-\cos x}\] \[=\log \,\,2\,.\underset{x\to 0}{\mathop{\lim }}\,\,\,\frac{{{x}^{2}}}{2\,\,{{\sin }^{2}}\frac{x}{2}}=(\log \,\,2)\,.\,2=2\log 2=\log 4\].
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