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Question: $\lim_{x \to 2} \frac{\sqrt{1 - \cos\{2(x-2)\}}}{x-2}$...

lim⁡x→21−cos⁡{2(x−2)}x−2\lim_{x \to 2} \frac{\sqrt{1 - \cos\{2(x-2)\}}}{x-2}

Answer

The limit does not exist.

Explanation

Solution

Let y=x−2y = x-2. As x→2x \to 2, y→0y \to 0. The limit becomes: lim⁡y→01−cos⁡(2y)y\lim_{y \to 0} \frac{\sqrt{1 - \cos(2y)}}{y} Using the identity 1−cos⁡(2y)=2sin⁡2(y)1 - \cos(2y) = 2\sin^2(y), the expression is: lim⁡y→02sin⁡2(y)y=lim⁡y→02∣sin⁡(y)∣y\lim_{y \to 0} \frac{\sqrt{2\sin^2(y)}}{y} = \lim_{y \to 0} \frac{\sqrt{2}|\sin(y)|}{y} We evaluate the one-sided limits: For the right-hand limit (y→0+y \to 0^+): lim⁡y→0+2sin⁡(y)y=2lim⁡y→0+sin⁡(y)y=2×1=2\lim_{y \to 0^+} \frac{\sqrt{2}\sin(y)}{y} = \sqrt{2} \lim_{y \to 0^+} \frac{\sin(y)}{y} = \sqrt{2} \times 1 = \sqrt{2} For the left-hand limit (y→0−y \to 0^-): lim⁡y→0−2(−sin⁡(y))y=−2lim⁡y→0−sin⁡(y)y=−2×1=−2\lim_{y \to 0^-} \frac{\sqrt{2}(-\sin(y))}{y} = -\sqrt{2} \lim_{y \to 0^-} \frac{\sin(y)}{y} = -\sqrt{2} \times 1 = -\sqrt{2} Since the right-hand limit (2\sqrt{2}) and the left-hand limit (−2-\sqrt{2}) are not equal, the limit does not exist.