Solution :
\(\displaystyle{\lim_{x \to 0}}\) \(xln(1+2tanx)\over 1-cosx\)
= \(\displaystyle{\lim_{x \to 0}}\) \(xln(1+2tanx)\over {1-cosx\over x^2}.x^2\).\(2tanx\over 2tanx\)
= 4
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