Solution :
\(\displaystyle{\lim_{x \to 0}}\) \(2(sin(2+x)-sin2)+xsin(2+x)\over x\)
= \(\displaystyle{\lim_{x \to 0}}\)(\(2.2.cos(2+{x\over 2})sin{x\over 2}\over x\) + sin(2+x))
= \(\displaystyle{\lim_{x \to 0}}\)\(2cos(2+{x\over 2})sin{x\over 2}\over {x\over 2}\) + \(\displaystyle{\lim_{x \to 0}}\)sin(2+x)
= 2cos2 + sin2
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