A) \[sin\text{ }2\]
B) \[cos\text{ }2\]
C) 1
D) \[2\text{ }cos\text{ }2+sin\text{ }2\]
E) \[2\text{ }sin\text{ }2+cos\text{ }2\]
Correct Answer: D
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{(2+x)\sin (2+x)-2\sin 2}{x}\] \[\left( \frac{0}{0}form \right)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin (2+x)+(2+x)\cos (2+x)}{1}\] \[=\sin 2+2\cos 2\]
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