Key Concepts and Formulas

• Trigonometric Identity: The fundamental identity for simplifying expressions involving $1 - \cos(2\theta)$ is $1 - \cos(2\theta) = 2\sin^2(\theta)$.
• Absolute Value: The square root of a squared term is its absolute value: $\sqrt{a^2} = |a|$.
• Limit Definition: A limit exists at a point if and only if the left-hand limit (LHL) and the right-hand limit (RHL) exist and are equal.
• Standard Limit: The standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ will be used.

Step-by-Step Solution

Step 1: Simplify the expression using trigonometric identities. We are given the limit: $L = \mathop {\lim }\limits_{x \to 2} \left( {{{\sqrt {1 - \cos \left\{ {2(x - 2)} \right\}} } \over {x - 2}}} \right)$ Let $\theta = x - 2$. As $x \to 2$, we have $\theta \to 0$. The expression inside the limit becomes: $\sqrt{1 - \cos\{2\theta\}}$ Using the identity $1 - \cos(2\alpha) = 2\sin^2(\alpha)$, with $\alpha = \theta$, we get: $\sqrt{1 - \cos\{2\theta\}} = \sqrt{2\sin^2(\theta)}$ Applying the property $\sqrt{a^2} = |a|$, we have: $\sqrt{2\sin^2(\theta)} = \sqrt{2}|\sin(\theta)|$ Substituting back $\theta = x - 2$, the expression becomes: $\sqrt{2}|\sin(x - 2)|$ So, the limit can be rewritten as: $L = \mathop {\lim }\limits_{x \to 2} \left( {{\sqrt 2 |\sin(x - 2)|} \over {x - 2}} \right)$

Step 2: Evaluate the Left-Hand Limit (LHL). To evaluate the limit as $x \to 2^-$, we consider values of $x$ slightly less than 2. This means $x - 2$ is a small negative number. Let $x \to 2^-$. Then $x - 2 \to 0^-$. For small negative values of $\phi$, $\sin(\phi)$ is also negative. Therefore, for $x - 2 \to 0^-$, $\sin(x - 2)$ is negative. This implies that $|\sin(x - 2)| = -\sin(x - 2)$ when $x - 2 < 0$. So, the LHL is: $LHL = \mathop {\lim }\limits_{x \to {2^- }} \left( {{\sqrt 2 (-\sin(x - 2))} \over {x - 2}} \right)$ $LHL = -\sqrt 2 \mathop {\lim }\limits_{x \to {2^- }} \left( {{\sin(x - 2)} \over {x - 2}} \right)$ Let $y = x - 2$. As $x \to 2^-$, $y \to 0^-$. The limit becomes: $LHL = -\sqrt 2 \mathop {\lim }\limits_{y \to {0^- }} \left( {{\sin(y)} \over {y}} \right)$ Using the standard limit $\mathop {\lim }\limits_{y \to 0} \frac{\sin y}{y} = 1$, the LHL is: $LHL = -\sqrt 2 \times 1 = -\sqrt 2$

Step 3: Evaluate the Right-Hand Limit (RHL). To evaluate the limit as $x \to 2^+$, we consider values of $x$ slightly greater than 2. This means $x - 2$ is a small positive number. Let $x \to 2^+$. Then $x - 2 \to 0^+$. For small positive values of $\phi$, $\sin(\phi)$ is also positive. Therefore, for $x - 2 \to 0^+$, $\sin(x - 2)$ is positive. This implies that $|\sin(x - 2)| = \sin(x - 2)$ when $x - 2 > 0$. So, the RHL is: $RHL = \mathop {\lim }\limits_{x \to {2^+ }} \left( {{\sqrt 2 \sin(x - 2)} \over {x - 2}} \right)$ Let $y = x - 2$. As $x \to 2^+$, $y \to 0^+$. The limit becomes: $RHL = \sqrt 2 \mathop {\lim }\limits_{y \to {0^+ }} \left( {{\sin(y)} \over {y}} \right)$ Using the standard limit $\mathop {\lim }\limits_{y \to 0} \frac{\sin y}{y} = 1$, the RHL is: $RHL = \sqrt 2 \times 1 = \sqrt 2$

Step 4: Compare LHL and RHL to determine if the limit exists. We found that $LHL = -\sqrt 2$ and $RHL = \sqrt 2$. Since $LHL \ne RHL$, the limit of the given function as $x \to 2$ does not exist.

Common Mistakes & Tips

• Forgetting the absolute value: When simplifying $\sqrt{\sin^2(\theta)}$, it is crucial to remember that $\sqrt{a^2} = |a|$, not just $a$. This absolute value is what leads to different LHL and RHL.
• Incorrect sign for $\sin(x-2)$: Be careful about the sign of $\sin(x-2)$ in different neighborhoods of $x=2$. For $x<2$, $x-2<0$, and $\sin(x-2)<0$. For $x>2$, $x-2>0$, and $\sin(x-2)>0$.
• Misapplication of the standard limit: Ensure that the argument of the sine function and the denominator are approaching zero together, and that the expression is in the form $\frac{\sin(\theta)}{\theta}$.

Summary

The problem requires evaluating a limit involving a square root and a trigonometric function. We first simplified the expression using the identity $1 - \cos(2\theta) = 2\sin^2(\theta)$ and the property of square roots $\sqrt{a^2} = |a|$. This resulted in the expression $\frac{\sqrt{2}|\sin(x-2)|}{x-2}$. To determine if the limit exists, we calculated the left-hand limit and the right-hand limit separately. For the LHL, since $x \to 2^-$, $x-2$ is negative, making $|\sin(x-2)| = -\sin(x-2)$. For the RHL, since $x \to 2^+$, $x-2$ is positive, making $|\sin(x-2)| = \sin(x-2)$. Both limits were evaluated using the standard limit $\mathop {\lim }\limits_{y \to 0} \frac{\sin y}{y} = 1$. The LHL was found to be $-\sqrt{2}$ and the RHL was found to be $\sqrt{2}$. Since the LHL and RHL are not equal, the overall limit does not exist.

The final answer is (D).