Key Concepts and Formulas

• Trigonometric Identities:
  • $\tan(\frac{\pi}{4} - \theta) = \frac{\tan(\frac{\pi}{4}) - \tan(\theta)}{1 + \tan(\frac{\pi}{4})\tan(\theta)} = \frac{1 - \tan(\theta)}{1 + \tan(\theta)}$
  • $1 - \sin x = 1 - \cos(\frac{\pi}{2} - x)$
  • $1 - \cos y = 2 \sin^2(\frac{y}{2})$
  • $\sin(\frac{\pi}{2} - x) = \cos x$
• Standard Limits:
  • $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$
  • $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$
• Limit Properties: The limit of a product is the product of the limits, provided the individual limits exist.

Step-by-Step Solution

Let the given limit be $L$. $L = \mathop {\lim }\limits_{x \to {\pi \over 2}} {{\left[ {1 - \tan \left( {{x \over 2}} \right)} \right]\left[ {1 - \sin x} \right]} \over {\left[ {1 + \tan \left( {{x \over 2}} \right)} \right]{{\left[ {\pi - 2x} \right]}^3}}}$

Step 1: Rewrite the tangent term using a trigonometric identity. We observe that the term $\frac{1 - \tan(x/2)}{1 + \tan(x/2)}$ can be rewritten. Let $\theta = x/2$. Then the expression is $\frac{1 - \tan \theta}{1 + \tan \theta}$. We know the identity $\tan(\frac{\pi}{4} - \theta) = \frac{\tan(\frac{\pi}{4}) - \tan(\theta)}{1 + \tan(\frac{\pi}{4})\tan(\theta)} = \frac{1 - \tan(\theta)}{1 + \tan(\theta)}$. Substituting $\theta = x/2$, we get $\frac{1 - \tan(x/2)}{1 + \tan(x/2)} = \tan(\frac{\pi}{4} - \frac{x}{2})$. So, the limit becomes: $L = \mathop {\lim }\limits_{x \to {\pi \over 2}} {{\tan \left( {\frac{\pi}{4} - \frac{x}{2}} \right)\left[ {1 - \sin x} \right]} \over {{{\left[ {\pi - 2x} \right]}^3}}}$

Step 2: Introduce a substitution to simplify the limit as $x \to \pi/2$. Let $x = \frac{\pi}{2} + y$. As $x \to \frac{\pi}{2}$, we have $y \to 0$. We need to express the terms in the limit in terms of $y$:

• $\frac{\pi}{4} - \frac{x}{2} = \frac{\pi}{4} - \frac{1}{2}(\frac{\pi}{2} + y) = \frac{\pi}{4} - \frac{\pi}{4} - \frac{y}{2} = -\frac{y}{2}$.
• $1 - \sin x = 1 - \sin(\frac{\pi}{2} + y) = 1 - \cos y$.
• $\pi - 2x = \pi - 2(\frac{\pi}{2} + y) = \pi - \pi - 2y = -2y$. Therefore, $(\pi - 2x)^3 = (-2y)^3 = -8y^3$.

Substituting these into the limit expression: $L = \mathop {\lim }\limits_{y \to 0} {{\tan \left( {-\frac{y}{2}} \right)\left( {1 - \cos y} \right)} \over {{{\left( {-2y} \right)}^3}}}$

Step 3: Apply trigonometric identities to simplify the numerator. We use the identity $1 - \cos y = 2 \sin^2(\frac{y}{2})$ and the property $\tan(-\theta) = -\tan(\theta)$. $L = \mathop {\lim }\limits_{y \to 0} {{-\tan \left( {\frac{y}{2}} \right)\left( {2 \sin^2\left(\frac{y}{2}\right)} \right)} \over {-8y^3}}$

Step 4: Rearrange the terms and use standard limit forms. We can cancel the negative signs and rearrange the terms to isolate the standard limit forms. $L = \mathop {\lim }\limits_{y \to 0} {{2 \tan \left( {\frac{y}{2}} \right) \sin^2\left(\frac{y}{2}\right)} \over {8y^3}}$ $L = \mathop {\lim }\limits_{y \to 0} {{1} \over {4}} \cdot {{\tan \left( {\frac{y}{2}} \right)} \over {y}} \cdot {{\sin^2\left(\frac{y}{2}\right)} \over {y^2}}$ To match the standard limits $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$ and $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, we need to adjust the denominators.

Let's rewrite the expression as: $L = \mathop {\lim }\limits_{y \to 0} {{1} \over {4}} \cdot \left( {{\tan \left( {\frac{y}{2}} \right)} \over {\frac{y}{2}}} \cdot \frac{1}{2} \right) \cdot \left( {{{\sin \left(\frac{y}{2}\right)} \over {\frac{y}{2}}}} \cdot \frac{1}{2} \right)^2$ $L = \mathop {\lim }\limits_{y \to 0} {{1} \over {4}} \cdot \left( {{\tan \left( {\frac{y}{2}} \right)} \over {\frac{y}{2}}} \right) \cdot \frac{1}{2} \cdot \left( {{{\sin \left(\frac{y}{2}\right)} \over {\frac{y}{2}}}} \right)^2 \cdot \frac{1}{4}$ $L = \mathop {\lim }\limits_{y \to 0} {{1} \over {32}} \cdot \left( {{\tan \left( {\frac{y}{2}} \right)} \over {\frac{y}{2}}} \right) \cdot \left( {{{\sin \left(\frac{y}{2}\right)} \over {\frac{y}{2}}}} \right)^2$

Step 5: Evaluate the limit using the standard forms. As $y \to 0$, $\frac{y}{2} \to 0$. Therefore, we can use the standard limits: $\lim_{y \to 0} \frac{\tan(y/2)}{y/2} = 1$ $\lim_{y \to 0} \frac{\sin(y/2)}{y/2} = 1$

Substituting these values: $L = {1 \over 32} \cdot (1) \cdot (1)^2$ $L = {1 \over 32}$

Let's re-examine the provided solution and the problem statement. The provided solution states the answer is $1/32$, which corresponds to option (D). However, the problem states the correct answer is (A) $\infty$. This indicates a discrepancy. Let's re-evaluate the problem carefully to see if there was a mistake in the simplification or if the provided "correct answer" is indeed incorrect.

Let's re-examine Step 4 and 5. $L = \mathop {\lim }\limits_{y \to 0} {{2 \tan \left( {\frac{y}{2}} \right) \sin^2\left(\frac{y}{2}\right)} \over {8y^3}}$ $L = \mathop {\lim }\limits_{y \to 0} {{1} \over {4}} \cdot \frac{\tan(y/2)}{y/2} \cdot \frac{1}{2} \cdot \frac{\sin^2(y/2)}{y^2}$ $L = \mathop {\lim }\limits_{y \to 0} {{1} \over {8}} \cdot \frac{\tan(y/2)}{y/2} \cdot \frac{\sin^2(y/2)}{y^2}$ We need $y^2$ in the denominator for $\sin^2(y/2)$. We have $y^3$ in the denominator. Let's group the terms differently: $L = \mathop {\lim }\limits_{y \to 0} {{1} \over {8}} \cdot \tan\left(\frac{y}{2}\right) \cdot \frac{\sin^2\left(\frac{y}{2}\right)}{y^3}$ $L = \mathop {\lim }\limits_{y \to 0} {{1} \over {8}} \cdot \frac{\tan\left(\frac{y}{2}\right)}{\frac{y}{2}} \cdot \frac{1}{2} \cdot \frac{\sin^2\left(\frac{y}{2}\right)}{y^2}$ $L = \mathop {\lim }\limits_{y \to 0} {{1} \over {16}} \cdot \frac{\tan\left(\frac{y}{2}\right)}{\frac{y}{2}} \cdot \frac{\sin^2\left(\frac{y}{2}\right)}{y^2}$ We still need to adjust the $\sin^2(y/2)$ term. $L = \mathop {\lim }\limits_{y \to 0} {{1} \over {16}} \cdot \left( \frac{\tan(y/2)}{y/2} \right) \cdot \left( \frac{\sin(y/2)}{y/2} \right)^2 \cdot \left(\frac{y/2}{y}\right)^2$ $L = \mathop {\lim }\limits_{y \to 0} {{1} \over {16}} \cdot \left( \frac{\tan(y/2)}{y/2} \right) \cdot \left( \frac{\sin(y/2)}{y/2} \right)^2 \cdot \left(\frac{1}{2}\right)^2$ $L = \mathop {\lim }\limits_{y \to 0} {{1} \over {16}} \cdot \left( \frac{\tan(y/2)}{y/2} \right) \cdot \left( \frac{\sin(y/2)}{y/2} \right)^2 \cdot \frac{1}{4}$ $L = {1 \over 64} \cdot (1) \cdot (1)^2 = {1 \over 64}$

There seems to be a mistake in the original solution's simplification. Let's re-evaluate the original solution's steps carefully. The original solution has: $= \mathop {\lim }\limits_{y \to 0} {{ - \tan {y \over 2}2{{\sin }^2}{y \over 2}} \over {\left( { - 8} \right).{{{y^3}} \over 8}.8}}$ The denominator here seems to be simplified incorrectly. It says $(-8) \cdot (y^3/8) \cdot 8$. This seems to be a typo. It should be $(-2y)^3 = -8y^3$.

Let's follow the original solution's structure but correct the simplification. $L = \mathop {\lim }\limits_{y \to 0} {{\tan \left( {-\frac{y}{2}} \right)\left( {1 - \cos y} \right)} \over {{{\left( {-2y} \right)}^3}}}$ $L = \mathop {\lim }\limits_{y \to 0} {{-\tan \left( {\frac{y}{2}} \right) \cdot 2{{\sin }^2}\left(\frac{y}{2}\right)} \over {-8y^3}}$ $L = \mathop {\lim }\limits_{y \to 0} {{\tan \left( {\frac{y}{2}} \right) \cdot 2{{\sin }^2}\left(\frac{y}{2}\right)} \over {8y^3}}$ $L = \mathop {\lim }\limits_{y \to 0} {{1} \over {4}} \cdot \frac{\tan(y/2)}{y} \cdot \frac{\sin^2(y/2)}{y^2}$ We need to get $y/2$ in the denominator for $\tan(y/2)$ and $y/2$ for $\sin(y/2)$. $L = \mathop {\lim }\limits_{y \to 0} {{1} \over {4}} \cdot \frac{\tan(y/2)}{y/2} \cdot \frac{1}{2} \cdot \frac{\sin^2(y/2)}{y^2}$ $L = \mathop {\lim }\limits_{y \to 0} {{1} \over {8}} \cdot \frac{\tan(y/2)}{y/2} \cdot \frac{\sin^2(y/2)}{y^2}$ Now, for $\sin^2(y/2)$, we need $(y/2)^2$ in the denominator. $L = \mathop {\lim }\limits_{y \to 0} {{1} \over {8}} \cdot \frac{\tan(y/2)}{y/2} \cdot \left( \frac{\sin(y/2)}{y/2} \right)^2 \cdot \left(\frac{y/2}{y}\right)^2$ $L = \mathop {\lim }\limits_{y \to 0} {{1} \over {8}} \cdot \frac{\tan(y/2)}{y/2} \cdot \left( \frac{\sin(y/2)}{y/2} \right)^2 \cdot \left(\frac{1}{2}\right)^2$ $L = \mathop {\lim }\limits_{y \to 0} {{1} \over {8}} \cdot \frac{\tan(y/2)}{y/2} \cdot \left( \frac{\sin(y/2)}{y/2} \right)^2 \cdot \frac{1}{4}$ $L = {1 \over 32} \cdot (1) \cdot (1)^2 = {1 \over 32}$

The original solution's final answer of $1/32$ appears to be correct based on this detailed re-derivation. However, the problem statement says the correct answer is (A) $\infty$. This suggests there might be an error in the problem statement or the provided correct answer. Let's assume for a moment that the provided correct answer (A) $\infty$ is indeed correct and try to find a reason.

Let's re-examine the limit expression and the behavior of terms as $x \to \pi/2$. $L = \mathop {\lim }\limits_{x \to {\pi \over 2}} {{\left[ {1 - \tan \left( {{x \over 2}} \right)} \right]\left[ {1 - \sin x} \right]} \over {\left[ {1 + \tan \left( {{x \over 2}} \right)} \right]{{\left[ {\pi - 2x} \right]}^3}}}$ As $x \to \pi/2$:

• $x/2 \to \pi/4$.
• $\tan(x/2) \to \tan(\pi/4) = 1$.
• $1 - \tan(x/2) \to 1 - 1 = 0$.
• $1 + \tan(x/2) \to 1 + 1 = 2$.
• $\sin x \to \sin(\pi/2) = 1$, so $1 - \sin x \to 1 - 1 = 0$.
• $\pi - 2x \to \pi - 2(\pi/2) = 0$.

We have an indeterminate form of type $\frac{0 \cdot 0}{2 \cdot 0^3}$, which is $\frac{0}{0}$.

Let's use Taylor series expansions around $x = \pi/2$. Let $x = \frac{\pi}{2} + h$, so $h \to 0$.

• $\frac{x}{2} = \frac{\pi}{4} + \frac{h}{2}$.

• $\tan(\frac{x}{2}) = \tan(\frac{\pi}{4} + \frac{h}{2})$. Using $\tan(A+B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$: $\tan(\frac{\pi}{4} + \frac{h}{2}) = \frac{\tan(\pi/4) + \tan(h/2)}{1 - \tan(\pi/4)\tan(h/2)} = \frac{1 + \tan(h/2)}{1 - \tan(h/2)}$. For small $h$, $\tan(h/2) \approx h/2$. So, $\tan(\frac{x}{2}) \approx \frac{1 + h/2}{1 - h/2} \approx (1 + h/2)(1 + h/2) \approx 1 + h$. Then $1 - \tan(\frac{x}{2}) \approx 1 - (1 + h) = -h$. And $1 + \tan(\frac{x}{2}) \approx 1 + (1 + h) = 2 + h \approx 2$.

• $\sin x = \sin(\frac{\pi}{2} + h) = \cos h$. For small $h$, $\cos h \approx 1 - \frac{h^2}{2}$. So, $1 - \sin x \approx 1 - (1 - \frac{h^2}{2}) = \frac{h^2}{2}$.

• $\pi - 2x = \pi - 2(\frac{\pi}{2} + h) = \pi - \pi - 2h = -2h$. So, $(\pi - 2x)^3 = (-2h)^3 = -8h^3$.

Now substitute these approximations into the limit: $L \approx \mathop {\lim }\limits_{h \to 0} \frac{(-h) (\frac{h^2}{2})}{(2) (-8h^3)}$ $L \approx \mathop {\lim }\limits_{h \to 0} \frac{-\frac{1}{2}h^3}{-16h^3}$ $L \approx \frac{-1/2}{-16} = \frac{1}{32}$

The Taylor expansion method also yields $1/32$. This strongly suggests that the provided "correct answer" of $\infty$ is incorrect for this problem. Assuming the problem is stated correctly and the options are correct, and given that our derivation consistently leads to $1/32$, we will proceed with this answer.

Let's review the original solution again to see if there's any subtle error that might lead to infinity. The original solution: $\mathop {\lim }\limits_{x \to {\pi \over 2}} {{\tan \left( {{\pi \over 4} - {x \over 2}} \right).\left( {1 - \sin x} \right)} \over {{{\left( {\pi - 2x} \right)}^3}}}$ Let $x = {\pi \over 2} + y$, so $y \to 0$. $= \mathop {\lim }\limits_{y \to 0} {{\tan \left( { - {y \over 2}} \right).\left( {1 - \cos \,y} \right)} \over {{{\left( { - 2y} \right)}^3}}}$ $= \mathop {\lim }\limits_{y \to 0} {{ - \tan {y \over 2} \cdot 2{{\sin }^2}{y \over 2}} \over { - 8{y^3}}}$ $= \mathop {\lim }\limits_{y \to 0} {{ \tan {y \over 2} \cdot 2{{\sin }^2}{y \over 2}} \over { 8{y^3}}}$ $= \mathop {\lim }\limits_{y \to 0} {{1 \over 4} \cdot \frac{\tan(y/2)}{y} \cdot \frac{\sin^2(y/2)}{y^2}}$ This is where the original solution has a slight simplification issue. Let's correct it. $= \mathop {\lim }\limits_{y \to 0} {{1 \over 4} \cdot \frac{\tan(y/2)}{y/2} \cdot \frac{1}{2} \cdot \left(\frac{\sin(y/2)}{y/2}\right)^2 \cdot \left(\frac{y/2}{y}\right)^2}$ $= \mathop {\lim }\limits_{y \to 0} {{1 \over 4} \cdot \frac{\tan(y/2)}{y/2} \cdot \frac{1}{2} \cdot \left(\frac{\sin(y/2)}{y/2}\right)^2 \cdot \left(\frac{1}{2}\right)^2}$ $= \mathop {\lim }\limits_{y \to 0} {{1 \over 4} \cdot \frac{1}{2} \cdot \frac{1}{4} \cdot \frac{\tan(y/2)}{y/2} \cdot \left(\frac{\sin(y/2)}{y/2}\right)^2}$ $= {1 \over 32} \cdot (1) \cdot (1)^2 = {1 \over 32}$

The original solution's calculation of $1/32$ is correct. If the provided correct answer is indeed $\infty$, then there must be a misunderstanding of the question or a typo in the question itself. Given the consistency of our derivations, we will proceed with the result $1/32$.

Step 1: Rewrite the tangent term. Using the identity $\frac{1 - \tan \theta}{1 + \tan \theta} = \tan(\frac{\pi}{4} - \theta)$, we rewrite the expression as: $L = \mathop {\lim }\limits_{x \to {\pi \over 2}} {{\tan \left( {\frac{\pi}{4} - \frac{x}{2}} \right)\left[ {1 - \sin x} \right]} \over {{{\left[ {\pi - 2x} \right]}^3}}}$

Step 2: Perform a substitution. Let $x = \frac{\pi}{2} + y$. As $x \to \frac{\pi}{2}$, $y \to 0$. The expression transforms to: $L = \mathop {\lim }\limits_{y \to 0} {{\tan \left( {\frac{\pi}{4} - \frac{1}{2}(\frac{\pi}{2} + y)} \right)\left[ {1 - \sin(\frac{\pi}{2} + y)} \right]} \over {{{\left[ {\pi - 2(\frac{\pi}{2} + y)} \right]}^3}}}$ $L = \mathop {\lim }\limits_{y \to 0} {{\tan \left( {-\frac{y}{2}} \right)\left[ {1 - \cos y} \right]} \over {{{\left[ {-2y} \right]}^3}}}$

Step 3: Apply trigonometric identities. Using $\tan(-\theta) = -\tan \theta$ and $1 - \cos y = 2 \sin^2(\frac{y}{2})$: $L = \mathop {\lim }\limits_{y \to 0} {{-\tan \left( {\frac{y}{2}} \right) \cdot 2{{\sin }^2}\left(\frac{y}{2}\right)} \over {-8y^3}}$ $L = \mathop {\lim }\limits_{y \to 0} {{\tan \left( {\frac{y}{2}} \right) \cdot 2{{\sin }^2}\left(\frac{y}{2}\right)} \over {8y^3}}$

Step 4: Rearrange and use standard limit forms. $L = \mathop {\lim }\limits_{y \to 0} {{1} \over {4}} \cdot \frac{\tan(y/2)}{y} \cdot \frac{\sin^2(y/2)}{y^2}$ To use the standard limits $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$ and $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, we adjust the denominators: $L = \mathop {\lim }\limits_{y \to 0} {{1} \over {4}} \cdot \left( \frac{\tan(y/2)}{y/2} \cdot \frac{1}{2} \right) \cdot \left( \frac{\sin(y/2)}{y/2} \cdot \frac{1}{2} \right)^2$ $L = \mathop {\lim }\limits_{y \to 0} {{1} \over {4}} \cdot \frac{1}{2} \cdot \left( \frac{\tan(y/2)}{y/2} \right) \cdot \frac{1}{4} \cdot \left( \frac{\sin(y/2)}{y/2} \right)^2$ $L = \mathop {\lim }\limits_{y \to 0} {{1} \over {32}} \cdot \frac{\tan(y/2)}{y/2} \cdot \left( \frac{\sin(y/2)}{y/2} \right)^2$

Step 5: Evaluate the limit. As $y \to 0$, $\frac{y}{2} \to 0$. Using the standard limits: $L = {1 \over 32} \cdot (1) \cdot (1)^2 = {1 \over 32}$

Common Mistakes & Tips

• Algebraic Errors: Be extremely careful with signs and powers when simplifying trigonometric expressions and denominators, especially after substitution.
• Incorrect Standard Limit Application: Ensure that the argument of the trigonometric function in the numerator matches the argument in the denominator for standard limits like $\frac{\tan \theta}{\theta}$ and $\frac{\sin \theta}{\theta}$. You might need to multiply and divide by constants.
• Misinterpreting Indeterminate Forms: When faced with $\frac{0}{0}$, it's a signal to use algebraic manipulation, trigonometric identities, or Taylor series to resolve the indeterminacy.

Summary

The problem involves evaluating a limit of a trigonometric function. By rewriting the expression using the tangent subtraction formula, performing a substitution to simplify the limit as the variable approaches the limit point, and then applying standard trigonometric identities and standard limit forms, we were able to simplify the expression. Each step was carefully managed to ensure accuracy in algebraic and trigonometric manipulations. The final evaluation using the standard limits $\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$ and $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ leads to the result $1/32$. There appears to be a discrepancy with the provided correct answer.

Final Answer

The final answer is $\boxed{{1 \over 32}}$. This corresponds to option (D).