Key Concepts and Formulas

• Trigonometric Identities:
  • Angle Addition Formula for Sine: $\sin(A - B) = \sin A \cos B - \cos A \sin B$
  • Angle Addition Formula for Cosine: $\cos(A + B) = \cos A \cos B - \sin A \sin B$
  • Special Angle Values: $\sin(\pi/6) = 1/2$, $\cos(\pi/6) = \sqrt{3}/2$
• Standard Limits: $\mathop {\lim }\limits_{x \to 0} \frac{\sin x}{x} = 1$
• Limit Properties:
  • $\mathop {\lim }\limits_{x \to c} [k \cdot f(x)] = k \cdot \mathop {\lim }\limits_{x \to c} f(x)$
  • $\mathop {\lim }\limits_{x \to c} [f(x) \cdot g(x)] = \mathop {\lim }\limits_{x \to c} f(x) \cdot \mathop {\lim }\limits_{x \to c} g(x)$ (if limits exist)
  • $\mathop {\lim }\limits_{x \to c} [f(x) / g(x)] = [\mathop {\lim }\limits_{x \to c} f(x)] / [\mathop {\lim }\limits_{x \to c} g(x)]$ (if limits exist and denominator limit is non-zero)

Step-by-Step Solution

Let the given limit be $L$. $L = \mathop {\lim }\limits_{h \to 0} 2\left\{ {{{\sqrt 3 \sin \left( {{\pi \over 6} + h} \right) - \cos \left( {{\pi \over 6} + h} \right)} \over {\sqrt 3 h\left( {\sqrt 3 \cosh - \sinh } \right)}}} \right\}$

Step 1: Manipulate the numerator to use trigonometric identities. We observe that the numerator, $\sqrt{3} \sin(\frac{\pi}{6} + h) - \cos(\frac{\pi}{6} + h)$, can be simplified by factoring out a 2. This will allow us to use the sine subtraction formula. $\sqrt{3} \sin \left( \frac{\pi}{6} + h \right) - \cos \left( \frac{\pi}{6} + h \right) = 2 \left( \frac{\sqrt{3}}{2} \sin \left( \frac{\pi}{6} + h \right) - \frac{1}{2} \cos \left( \frac{\pi}{6} + h \right) \right)$ Recall that $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$. Substituting these values, we get: $= 2 \left( \cos \frac{\pi}{6} \sin \left( \frac{\pi}{6} + h \right) - \sin \frac{\pi}{6} \cos \left( \frac{\pi}{6} + h \right) \right)$ Using the identity $\sin(A - B) = \sin A \cos B - \cos A \sin B$, with $A = \frac{\pi}{6} + h$ and $B = \frac{\pi}{6}$, we have: $= 2 \sin \left( \left( \frac{\pi}{6} + h \right) - \frac{\pi}{6} \right) = 2 \sin(h)$

Step 2: Manipulate the denominator to use trigonometric identities. The denominator is $\sqrt{3} h (\sqrt{3} \cosh - \sinh)$. We can factor out a 2 from the term $(\sqrt{3} \cosh - \sinh)$ to relate it to a cosine addition formula. $\sqrt{3} \cosh - \sinh = 2 \left( \frac{\sqrt{3}}{2} \cosh - \frac{1}{2} \sinh \right)$ Again, using $\cos(\frac{\pi}{6}) = \frac{\sqrt{3}}{2}$ and $\sin(\frac{\pi}{6}) = \frac{1}{2}$: $= 2 \left( \cos \frac{\pi}{6} \cosh - \sin \frac{\pi}{6} \sinh \right)$ This expression does not directly match the cosine addition formula $\cos(A+B) = \cos A \cos B - \sin A \sin B$. However, if we rewrite it as: $= 2 \left( \cos \frac{\pi}{6} \cosh + \sin \frac{\pi}{6} (-\sinh) \right)$ This is still not a direct match. Let's re-examine the original problem statement and the intended simplification. The term $\sqrt{3} \cosh - \sinh$ suggests a potential simplification related to hyperbolic functions, but the presence of $\pi/6$ strongly indicates trigonometric functions.

Let's re-evaluate the denominator's structure. The term $(\sqrt{3} \cosh - \sinh)$ seems unusual in the context of the trigonometric numerator. Let's assume there might be a typo or a specific intended form. However, given the provided solution path, it seems the denominator's hyperbolic part is meant to be simplified using a trigonometric identity. This is only possible if "cosh" and "sinh" are placeholders for trigonometric functions or if a specific transformation is applied.

Let's assume that the denominator's structure is meant to be related to the numerator's structure. If we consider the possibility of a typo and that the denominator should involve trigonometric functions, let's look at the provided solution's step: $\sqrt 3 h\left( {\sqrt 3 \cosh - \sinh } \right)$ The solution proceeds to rewrite this as: $2\sqrt 3 h\left( {{{\sqrt 3 } \over 2}\cosh - {1 \over 2}\sinh } \right)$ And then: $2\sqrt 3 h\left( {\cos {\pi \over 6}\cosh - \sin {\pi \over 6}\sinh } \right)$ This step suggests that $\cosh$ is being treated as $\cos(h)$ and $\sinh$ as $\sin(h)$. If we make this substitution, the denominator becomes: $\sqrt{3} h (\sqrt{3} \cos h - \sin h)$ This still doesn't align with the solution's next step, which suggests a form like $\cos(\frac{\pi}{6})\cos(h) - \sin(\frac{\pi}{6})\sin(h) = \cos(\frac{\pi}{6} + h)$.

Let's carefully re-examine the original solution provided. It seems to have made an implicit substitution or interpretation that is not explicitly stated. The original solution states: $\Rightarrow L = \mathop {\lim }\limits_{h \to 0} 2 \times 2\left\{ {{{{{\sqrt 3 } \over 2}\sin \left( {{\pi \over 6} + h} \right) - {1 \over 2}\cos \left( {{\pi \over 6} + h} \right)} \over {2\sqrt 3 h\left( {{{\sqrt 3 } \over 2}\cosh - {1 \over 2}\sinh } \right)}}} \right\}$ And then: $\Rightarrow L = \mathop {\lim }\limits_{h \to 0} 2 \times 2\left\{ {{{\cos {\pi \over 6}\sin \left( {{\pi \over 6} + h} \right) - \sin {\pi \over 6}\cos \left( {{\pi \over 6} + h} \right)} \over {2\sqrt 3 h\left( {\cos {\pi \over 6}\cosh - \sin {\pi \over 6}\sinh } \right)}}} \right\}$ The crucial leap happens here. The numerator is correctly identified as $2 \sin(h)$. However, the denominator's term $(\sqrt{3} \cosh - \sinh)$ is transformed into $2(\cos(\frac{\pi}{6})\cosh - \sin(\frac{\pi}{6})\sinh)$. The next step in the provided solution is: $\Rightarrow L = \mathop {\lim }\limits_{h \to 0} 2 \times 2\left\{ {{{\sin \left( {{\pi \over 6} + h - {\pi \over 6}} \right)} \over {2\sqrt 3 h\cos \left( {h + {\pi \over 6}} \right)}}} \right\}$ This implies that $\cos {\pi \over 6}\cosh - \sin {\pi \over 6}\sinh$ was somehow interpreted as $\cos(h + \frac{\pi}{6})$. This is mathematically incorrect unless $\cosh$ is replaced by $\cos(h)$ and $\sinh$ by $\sin(h)$, and then the identity $\cos A \cos B - \sin A \sin B = \cos(A+B)$ is applied with $A=h$ and $B=\pi/6$.

Given the constraints to reach the provided correct answer, we will proceed with the implicit assumption made in the original solution that the hyperbolic terms in the denominator are to be treated as trigonometric terms for the purpose of applying the identity. This is a common occurrence in JEE problems where a specific form is intended.

Let's assume $\cosh \equiv \cos h$ and $\sinh \equiv \sin h$ for the purpose of applying the identity. The denominator is: $\sqrt{3} h (\sqrt{3} \cosh - \sinh)$. Factoring out 2 from the bracket: $2\sqrt{3} h (\frac{\sqrt{3}}{2} \cosh - \frac{1}{2} \sinh)$. Substituting $\cos(\pi/6)$ for $\sqrt{3}/2$ and $\sin(\pi/6)$ for $1/2$: $2\sqrt{3} h (\cos(\pi/6) \cosh - \sin(\pi/6) \sinh)$. Now, applying the assumption $\cosh \equiv \cos h$ and $\sinh \equiv \sin h$: $2\sqrt{3} h (\cos(\pi/6) \cos h - \sin(\pi/6) \sin h)$. Using the identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$, with $A = \pi/6$ and $B = h$: $2\sqrt{3} h \cos(\pi/6 + h)$.

So, the expression for the limit becomes: $L = \mathop {\lim }\limits_{h \to 0} 2 \left\{ \frac{2 \sin(h)}{\sqrt{3} h \cdot 2\sqrt{3} h \cos(\pi/6 + h)} \right\}$ There seems to be a discrepancy in the powers of $\sqrt{3}$ and the factor of $h$ in the denominator based on this interpretation. Let's re-examine the original solution's structure.

The original solution has: $\Rightarrow L = \mathop {\lim }\limits_{h \to 0} 2 \times 2\left\{ {{{\sin \left( {{\pi \over 6} + h - {\pi \over 6}} \right)} \over {2\sqrt 3 h\cos \left( {h + {\pi \over 6}} \right)}}} \right\}$ Numerator: $2 \sin(h)$ (correct from Step 1). Denominator: $2\sqrt{3} h \cos(h + \pi/6)$.

This implies that the original term $\sqrt{3} h (\sqrt{3} \cosh - \sinh)$ was simplified to $2\sqrt{3} h \cos(h + \pi/6)$. This means $\sqrt{3} (\sqrt{3} \cosh - \sinh) = 2\sqrt{3} \cos(h + \pi/6)$. $\sqrt{3} \cosh - \sinh = 2 \cos(h + \pi/6)$. $\frac{\sqrt{3}}{2} \cosh - \frac{1}{2} \sinh = \cos(h + \pi/6)$. This step is problematic if we interpret $\cosh$ and $\sinh$ as hyperbolic functions.

However, if we strictly follow the provided solution's logic: Step 1 (as done above): Numerator simplification. $\sqrt 3 \sin \left( \frac{\pi}{6} + h \right) - \cos \left( \frac{\pi}{6} + h \right) = 2 \sin(h)$

Step 2: Denominator simplification based on the provided solution's structure. The solution factors out a 2 from the initial expression: $L = \mathop {\lim }\limits_{h \to 0} 2 \times 2\left\{ {{{\sqrt 3 \sin \left( {{\pi \over 6} + h} \right) - \cos \left( {{\pi \over 6} + h} \right)} \over {2\sqrt 3 h\left( {{{\sqrt 3 } \over 2}\cosh - {1 \over 2}\sinh } \right)}}} \right\}$ This means the original factor of 2 outside the limit was split, and a factor of 2 was introduced in the numerator and denominator. The term in the bracket is now: $\frac{\sqrt 3 \sin \left( {{\pi \over 6} + h} \right) - \cos \left( {{\pi \over 6} + h} \right)}{2\sqrt 3 h\left( {{{\sqrt 3 } \over 2}\cosh - {1 \over 2}\sinh } \right)}$ Using the result from Step 1 for the numerator, we get: $\frac{2 \sin(h)}{2\sqrt 3 h\left( {{{\sqrt 3 } \over 2}\cosh - {1 \over 2}\sinh } \right)}$ The solution then proceeds to use the trigonometric identity: $\frac{{{{\sqrt 3 } \over 2}\sin \left( {{\pi \over 6} + h} \right) - {1 \over 2}\cos \left( {{\pi \over 6} + h} \right)}}{2\sqrt 3 h\left( {{{\sqrt 3 } \over 2}\cosh - {1 \over 2}\sinh } \right)} = \frac{\cos {\pi \over 6}\sin \left( {{\pi \over 6} + h} \right) - \sin {\pi \over 6}\cos \left( {{\pi \over 6} + h} \right)}{2\sqrt 3 h\left( {\cos {\pi \over 6}\cosh - \sin {\pi \over 6}\sinh } \right)}$ This step correctly applies the trigonometric identity to the numerator. The denominator's expression $(\sqrt{3} \cosh - \sinh)$ is still present.

Step 3: Applying the trigonometric identity to the denominator's structure (as implied by the solution). The solution then states: $\Rightarrow L = \mathop {\lim }\limits_{h \to 0} 2 \times 2\left\{ {{{\sin \left( {{\pi \over 6} + h - {\pi \over 6}} \right)} \over {2\sqrt 3 h\cos \left( {h + {\pi \over 6}} \right)}}} \right\}$ This implies that the term $(\cos {\pi \over 6}\cosh - \sin {\pi \over 6}\sinh)$ in the denominator was replaced by $\cos(h + \pi/6)$. This replacement is valid if we assume $\cosh \equiv \cos h$ and $\sinh \equiv \sin h$, and then apply $\cos A \cos B - \sin A \sin B = \cos(A+B)$. So, the term in the denominator becomes: $2\sqrt{3} h \cos(h + \pi/6)$.

The expression inside the limit is now: $4 \times \frac{\sin(h)}{2\sqrt{3} h \cos(h + \pi/6)}$

Step 4: Simplify and evaluate the limit. $L = \mathop {\lim }\limits_{h \to 0} 4 \times \frac{\sin(h)}{2\sqrt{3} h \cos(h + \pi/6)}$ We can separate the constant factors and the standard limit: $L = \frac{4}{2\sqrt{3}} \mathop {\lim }\limits_{h \to 0} \frac{\sin(h)}{h} \times \mathop {\lim }\limits_{h \to 0} \frac{1}{\cos(h + \pi/6)}$ We know that $\mathop {\lim }\limits_{h \to 0} \frac{\sin(h)}{h} = 1$. And as $h \to 0$, $\cos(h + \pi/6) \to \cos(\pi/6) = \frac{\sqrt{3}}{2}$. So, the limit becomes: $L = \frac{4}{2\sqrt{3}} \times 1 \times \frac{1}{\cos(\pi/6)}$ $L = \frac{2}{\sqrt{3}} \times \frac{1}{\sqrt{3}/2}$ $L = \frac{2}{\sqrt{3}} \times \frac{2}{\sqrt{3}}$ $L = \frac{4}{3}$

Common Mistakes & Tips

• Misinterpreting Hyperbolic Functions: The presence of "cosh" and "sinh" can be confusing. In this specific problem, they are used in a way that hints at trigonometric identities. Always check if the context (like the presence of $\pi/6$) suggests a trigonometric simplification is intended.
• Algebraic Errors: Be meticulous with factoring constants and simplifying expressions, especially when dealing with square roots and fractions. A single misstep can lead to an incorrect final answer.
• Applying Limit Properties: Ensure that the conditions for applying limit properties (like the product or quotient rule) are met. For instance, the limit of the denominator must be non-zero when using the quotient rule.

Summary

The problem involves evaluating a limit of a complex trigonometric expression. The key to solving this problem lies in recognizing and applying trigonometric identities to simplify both the numerator and the denominator. The numerator is simplified using the sine subtraction formula after factoring out a 2. The denominator, while initially containing hyperbolic function notation, is treated as if it follows a trigonometric cosine addition pattern, allowing for further simplification. By carefully manipulating the expression, separating terms, and applying the standard limit $\mathop {\lim }\limits_{h \to 0} \frac{\sin h}{h} = 1$, the limit can be evaluated. The final result obtained through these steps is $4/3$.

The final answer is \boxed{4/3}. This corresponds to option (A).