A) doesn't exist
B) infinite
C) 0
D) 2
Correct Answer: D
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\frac{1-\cos 2x}{{{x}^{2}}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{2{{\sin }^{2}}x}{{{x}^{2}}}\] \[=2\underset{x\to 0}{\mathop{\lim }}\,{{\left( \frac{\sin x}{x} \right)}^{2}}=2\]You need to login to perform this action.
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