What is \lim_{x\rightarrow0}\frac{x}{\sqrt{1-\cos 4x}} equal to ?
- A. \frac{1}{2\sqrt{2}}
- B. -\frac{1}{2\sqrt{2}}
- C. \sqrt{2}
- D. Limit does <strong>NOT</strong> exist ✓
Correct Answer: D. Limit does <strong>NOT</strong> exist
Explanation
Using the identity 1-\cos 4x = 2\sin^2 2x, the denominator becomes \sqrt{2}|\sin 2x|. The limit is \lim_{x\rightarrow 0} \frac{x}{\sqrt{2}|\sin 2x|}. The Right Hand Limit (x \rightarrow 0^+) is \frac{1}{2\sqrt{2}} and the Left Hand Limit (x \rightarrow 0^-) is -\frac{1}{2\sqrt{2}}. Since LHL \neq RHL, the limit does not exist.
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