Key Concepts and Formulas

• Algebraic Manipulation for Limits: Techniques like rationalization are crucial for simplifying indeterminate forms of limits, especially those involving square roots.
• Standard Trigonometric Limits: The fundamental limit $\mathop {\lim }\limits_{u \to 0} \frac{\sin u}{u} = 1$ and its variations are essential for evaluating limits involving trigonometric functions.
• Double Angle and Half Angle Formulas: Knowledge of formulas like $\sin^2 x = \frac{1 - \cos 2x}{2}$ (or $1 - \cos x = 2\sin^2(x/2)$) and $\cos x = 2\cos^2(x/2) - 1$ helps simplify trigonometric expressions.

Step-by-Step Solution

Step 1: Identify the Indeterminate Form We are asked to evaluate the limit: $L = \mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }}$ As $x \to 0$, $\sin^2 x \to 0^2 = 0$. Also, as $x \to 0$, $\cos x \to \cos 0 = 1$. So, $\sqrt{1 + \cos x} \to \sqrt{1+1} = \sqrt{2}$. Therefore, the denominator $\sqrt{2} - \sqrt{1 + \cos x} \to \sqrt{2} - \sqrt{2} = 0$. This results in the indeterminate form $\frac{0}{0}$, indicating that further manipulation is required.

Step 2: Rationalize the Denominator To handle the square roots in the denominator, we multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{2} + \sqrt{1 + \cos x}$. $L = \mathop {\lim }\limits_{x \to 0} \left( {{{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }}} \right) \times \left( {{{\sqrt 2 + \sqrt {1 + \cos x} } \over {\sqrt 2 + \sqrt {1 + \cos x} }}} \right)$ $L = \mathop {\lim }\limits_{x \to 0} {{{{{\sin }^2}x} \left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)} \over {(\sqrt 2)^2 - (\sqrt {1 + \cos x})^2}}$ $L = \mathop {\lim }\limits_{x \to 0} {{{{{\sin }^2}x} \left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)} \over {2 - (1 + \cos x)}}$ $L = \mathop {\lim }\limits_{x \to 0} {{{{{\sin }^2}x} \left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)} \over {1 - \cos x}}$

Step 3: Apply Trigonometric Identities We can now use the identity $1 - \cos x = 2\sin^2(x/2)$. $L = \mathop {\lim }\limits_{x \to 0} {{{{{\sin }^2}x} \left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)} \over {2{{\sin }^2}\left( {{x \over 2}} \right)}}$ We also know that $\sin x = 2\sin(x/2)\cos(x/2)$. Therefore, $\sin^2 x = (2\sin(x/2)\cos(x/2))^2 = 4\sin^2(x/2)\cos^2(x/2)$. Substitute this into the expression: $L = \mathop {\lim }\limits_{x \to 0} {{{4{{\sin }^2}\left( {{x \over 2}} \right){{\cos }^2}\left( {{x \over 2}} \right)} \left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)} \over {2{{\sin }^2}\left( {{x \over 2}} \right)}}$

Step 4: Simplify and Use Standard Limits Cancel out the common term $2\sin^2(x/2)$ from the numerator and denominator. $L = \mathop {\lim }\limits_{x \to 0} {2{{\cos }^2}\left( {{x \over 2}} \right) \left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)}$ Now, we can directly substitute $x=0$ since the expression is no longer indeterminate. As $x \to 0$:

• $\cos(x/2) \to \cos(0) = 1$. So, $\cos^2(x/2) \to 1^2 = 1$.
• $\cos x \to \cos 0 = 1$. So, $\sqrt{1 + \cos x} \to \sqrt{1+1} = \sqrt{2}$.

Substituting these values: $L = 2 \times (1) \times (\sqrt{2} + \sqrt{2})$ $L = 2 \times (2\sqrt{2})$ $L = 4\sqrt{2}$

Wait! Reviewing the provided correct answer, it is $\sqrt{2}$. Let's re-examine the steps to find the error. The original solution provided leads to $4\sqrt{2}$, which contradicts the correct answer. Let's restart from Step 3 with a different simplification approach.

Revised Step 3: Apply Trigonometric Identities (Alternative Approach) We have: $L = \mathop {\lim }\limits_{x \to 0} {{{{{\sin }^2}x} \left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)} \over {1 - \cos x}}$ We know the standard limit $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$, which implies $\mathop {\lim }\limits_{x \to 0} \frac{\sin^2 x}{x^2} = 1$. We also know that $1 - \cos x = 2\sin^2(x/2)$. Let's rewrite the expression to utilize these standard forms.

$L = \mathop {\lim }\limits_{x \to 0} \left( \frac{\sin^2 x}{x^2} \right) \times \left( \frac{x^2}{1 - \cos x} \right) \times \left( \sqrt 2 + \sqrt {1 + \cos x} \right)$

Step 4: Evaluate Each Part of the Limit We evaluate each part separately:

1. $\mathop {\lim }\limits_{x \to 0} \frac{\sin^2 x}{x^2} = \left( \mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} \right)^2 = 1^2 = 1$.

2. $\mathop {\lim }\limits_{x \to 0} \frac{x^2}{1 - \cos x}$. We know $1 - \cos x = 2\sin^2(x/2)$. So, $\mathop {\lim }\limits_{x \to 0} \frac{x^2}{2\sin^2(x/2)}$. We can rewrite this as $\mathop {\lim }\limits_{x \to 0} \frac{1}{2} \left( \frac{x}{\sin(x/2)} \right)^2$. To use the standard limit $\frac{\sin u}{u} \to 1$, let $u = x/2$. Then $x = 2u$. $\mathop {\lim }\limits_{x \to 0} \frac{1}{2} \left( \frac{2u}{\sin u} \right)^2 = \mathop {\lim }\limits_{u \to 0} \frac{1}{2} \left( 2 \frac{u}{\sin u} \right)^2 = \frac{1}{2} \times (2 \times 1)^2 = \frac{1}{2} \times 4 = 2$

3. $\mathop {\lim }\limits_{x \to 0} \left( \sqrt 2 + \sqrt {1 + \cos x} \right)$. As $x \to 0$, $\cos x \to 1$. So, $\sqrt{1 + \cos x} \to \sqrt{1+1} = \sqrt{2}$. Thus, the limit is $\sqrt{2} + \sqrt{2} = 2\sqrt{2}$.

Step 5: Combine the Results Now, multiply the limits of the three parts: $L = \left( \mathop {\lim }\limits_{x \to 0} \frac{\sin^2 x}{x^2} \right) \times \left( \mathop {\lim }\limits_{x \to 0} \frac{x^2}{1 - \cos x} \right) \times \left( \mathop {\lim }\limits_{x \to 0} \left( \sqrt 2 + \sqrt {1 + \cos x} \right) \right)$ $L = 1 \times 2 \times 2\sqrt{2}$ $L = 4\sqrt{2}$

There appears to be a persistent discrepancy with the provided "Correct Answer". Let's re-examine the problem and the provided solution's steps more carefully, assuming the provided correct answer (A) $\sqrt{2}$ is indeed correct and the initial solution was flawed.

Let's go back to: $L = \mathop {\lim }\limits_{x \to 0} {{{{{\sin }^2}x} \left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)} \over {1 - \cos x}}$

We use $\sin^2 x = 1 - \cos^2 x = (1 - \cos x)(1 + \cos x)$. $L = \mathop {\lim }\limits_{x \to 0} {{{(1 - \cos x)(1 + \cos x)} \left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)} \over {1 - \cos x}}$ For $x \neq 0$ and $x$ close to 0, $1 - \cos x \neq 0$, so we can cancel the term $(1 - \cos x)$. $L = \mathop {\lim }\limits_{x \to 0} {(1 + \cos x) \left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)}$ Now, substitute $x=0$: $L = (1 + \cos 0) (\sqrt{2} + \sqrt{1 + \cos 0})$ $L = (1 + 1) (\sqrt{2} + \sqrt{1 + 1})$ $L = (2) (\sqrt{2} + \sqrt{2})$ $L = 2 (2\sqrt{2})$ $L = 4\sqrt{2}$

It seems the provided "Correct Answer" (A) $\sqrt{2}$ is incorrect based on standard limit evaluation techniques for the given problem. The derivation consistently leads to $4\sqrt{2}$. Assuming there might be a typo in the question or the provided answer, I will proceed with the derivation that leads to $4\sqrt{2}$, as it is mathematically sound.

Let's re-evaluate the provided solution's steps carefully. Current Solution's Step 4: $\mathop {\lim }\limits_{x \to 0} {1 \over 2}{\left( {{{\sin x} \over x}} \right)^2}{x^2}{\left( {{{{x \over 2}} \over {\sin {x \over 2}}}} \right)^2}{1 \over {{{{x^2}} \over 4}}}\left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)$ This step seems to have an error in the rearrangement of terms.

Let's trace the correct path to $4\sqrt{2}$ from the point after rationalization: $L = \mathop {\lim }\limits_{x \to 0} {{{{{\sin }^2}x} \left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)} \over {1 - \cos x}}$ Use $\sin^2 x = 1 - \cos^2 x = (1 - \cos x)(1 + \cos x)$. $L = \mathop {\lim }\limits_{x \to 0} {{{(1 - \cos x)(1 + \cos x)} \left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)} \over {1 - \cos x}}$ $L = \mathop {\lim }\limits_{x \to 0} {(1 + \cos x) \left( {\sqrt 2 + \sqrt {1 + \cos x} } \right)}$ Substitute $x=0$: $L = (1 + \cos 0) (\sqrt{2} + \sqrt{1 + \cos 0})$ $L = (1 + 1) (\sqrt{2} + \sqrt{1 + 1})$ $L = (2) (\sqrt{2} + \sqrt{2})$ $L = 2 (2\sqrt{2})$ $L = 4\sqrt{2}$

Given the discrepancy, and the instruction to arrive at the "Correct Answer" provided, there might be a misinterpretation of the question or a typo in the question/options/answer. However, if forced to choose from the options and assuming option (A) $\sqrt{2}$ is correct, a significant error in my derivation or the problem statement must exist.

Let's try to reverse-engineer how $\sqrt{2}$ might be obtained. This would imply a cancellation or simplification that is not apparent from standard methods.

Let's assume, for the sake of reaching the given answer, that there was a mistake in the rationalization or identity application. If the denominator was $\sqrt{2} - \sqrt{2-\cos x}$, then it would be different.

Let's reconsider the initial solution's final calculation: ${1 \over 2} \times 4 \times 2\sqrt 2$ = $4\sqrt 2$ . This confirms the initial solution's result is $4\sqrt{2}$, contradicting the stated correct answer. This further reinforces the belief that the provided "Correct Answer" is incorrect for the given problem.

Assuming the question and options are correct, and the provided "Correct Answer" (A) $\sqrt{2}$ is indeed the target, there must be a subtle error in my understanding or application of limits. However, all standard methods lead to $4\sqrt{2}$.

Let's proceed with the most robust derivation that yields $4\sqrt{2}$, acknowledging the conflict with the stated correct answer.

Revised Step-by-Step Solution

Step 1: Identify the Indeterminate Form We are asked to evaluate the limit: $L = \mathop {\lim }\limits_{x \to 0} {{{{\sin }^2}x} \over {\sqrt 2 - \sqrt {1 + \cos x} }}$ As $x \to 0$, $\sin^2 x \to 0$ and $\sqrt{1 + \cos x} \to \sqrt{1+1} = \sqrt{2}$. This results in the indeterminate form $\frac{0}{0}$.

Step 2: Rationalize the Denominator Multiply the numerator and the denominator by the conjugate of the denominator, $\sqrt{2} + \sqrt{1 + \cos x}$: $L = \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}x}}{{\sqrt 2 - \sqrt {1 + \cos x}}} \times \frac{{\sqrt 2 + \sqrt {1 + \cos x}}}{{\sqrt 2 + \sqrt {1 + \cos x}}}$ $L = \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}x (\sqrt 2 + \sqrt {1 + \cos x})}}{{2 - (1 + \cos x)}}$ $L = \mathop {\lim }\limits_{x \to 0} \frac{{{{\sin }^2}x (\sqrt 2 + \sqrt {1 + \cos x})}}{{1 - \cos x}}$

Step 3: Apply Trigonometric Identities Use the identity $\sin^2 x = 1 - \cos^2 x = (1 - \cos x)(1 + \cos x)$: $L = \mathop {\lim }\limits_{x \to 0} \frac{{(1 - \cos x)(1 + \cos x) (\sqrt 2 + \sqrt {1 + \cos x})}}{{1 - \cos x}}$ For $x \to 0$ and $x \neq 0$, $1 - \cos x \neq 0$, so we can cancel the term $(1 - \cos x)$: $L = \mathop {\lim }\limits_{x \to 0} {(1 + \cos x) (\sqrt 2 + \sqrt {1 + \cos x})}$

Step 4: Evaluate the Limit by Substitution Now, substitute $x=0$ into the simplified expression: $L = (1 + \cos 0) (\sqrt 2 + \sqrt {1 + \cos 0})$ $L = (1 + 1) (\sqrt 2 + \sqrt {1 + 1})$ $L = (2) (\sqrt 2 + \sqrt 2)$ $L = 2 (2\sqrt 2)$ $L = 4\sqrt 2$

Common Mistakes & Tips

• Algebraic Errors: Mistakes in rationalization or applying trigonometric identities are common. Double-check each algebraic step.
• Misapplication of Standard Limits: Ensure that the form $\frac{\sin u}{u}$ is correctly manipulated. For example, $\mathop {\lim }\limits_{x \to 0} \frac{\sin(x/2)}{x/2} = 1$, not $\mathop {\lim }\limits_{x \to 0} \frac{\sin(x/2)}{x} = 1/2$.
• Ignoring Indeterminate Forms: Do not substitute $x=0$ directly if the form is indeterminate. Always simplify first.

Summary

The problem involves evaluating a limit that results in the indeterminate form $\frac{0}{0}$. The strategy employed is to first rationalize the denominator to remove the square roots. Subsequently, trigonometric identities are used to simplify the expression, particularly by relating $\sin^2 x$ to $1 - \cos x$. After cancellation of common terms, the limit can be evaluated by direct substitution, yielding the result $4\sqrt{2}$. There is a discrepancy between this derived answer and the provided correct answer, suggesting a potential error in the problem statement or the given correct option. Based on rigorous mathematical derivation, the limit evaluates to $4\sqrt{2}$.

Final Answer The derived limit is $4\sqrt{2}$. However, this does not match any of the options if option (A) $\sqrt{2}$ is indeed the correct answer. If we strictly follow the provided correct answer (A), then there is an error in the problem statement or the intended solution path. Assuming the problem is stated correctly, the answer is $4\sqrt{2}$. If we are forced to match option (A) $\sqrt{2}$, it implies a fundamental misunderstanding of the problem or a severe typo. Given the instruction to match the correct answer, and the persistent derivation of $4\sqrt{2}$, I cannot definitively reach $\sqrt{2}$ through valid mathematical steps.

However, if we assume there was a typo in the denominator and it was meant to be $\sqrt{2+ \cos x}$ instead of $\sqrt{2-\sqrt{1+\cos x}}$, or some other modification that leads to $\sqrt{2}$, that would require a different problem.

Given the constraints, and the fact that the provided solution in the prompt leads to $4\sqrt{2}$, I must conclude that the "Correct Answer: A" is incorrect for this problem.

The final answer is $\boxed{4\sqrt{2}}$. If option (A) $\sqrt{2}$ must be the answer, the problem statement or the provided solution is flawed.