Key Concepts and Formulas

• Trigonometric Identities:
  • $\cot x = \frac{\cos x}{\sin x}$
  • $\cos(\frac{\pi}{2} - \theta) = \sin \theta$
  • $\sin(\frac{\pi}{2} - \theta) = \cos \theta$
  • $1 - \cos \theta = 2 \sin^2(\frac{\theta}{2})$
• Standard Limits:
  • $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$
  • $\mathop {\lim }\limits_{\theta \to 0} \frac{\tan \theta}{\theta} = 1$ (though not directly used, it's a related standard limit)
  • $\mathop {\lim }\limits_{\theta \to 0} \cos \theta = 1$
• L'Hôpital's Rule (optional but can be a check): If a limit results in an indeterminate form like $\frac{0}{0}$ or $\frac{\infty}{\infty}$, we can differentiate the numerator and denominator separately and then evaluate the limit.
• Substitution: Changing variables to simplify the limit expression, especially when the limit point is not 0.

Step-by-Step Solution

We want to evaluate the limit: $L = \mathop {\lim }\limits_{x \to {\pi \over 2}} {{\cot x - \cos x} \over {{{\left( {\pi - 2x} \right)}^3}}}$

Step 1: Rewrite the trigonometric terms and identify the indeterminate form. First, express $\cot x$ in terms of $\sin x$ and $\cos x$: $\cot x = \frac{\cos x}{\sin x}$ Substitute this into the expression: $L = \mathop {\lim }\limits_{x \to {\pi \over 2}} {{\frac{\cos x}{\sin x} - \cos x} \over {{{\left( {\pi - 2x} \right)}^3}}}$ Combine the terms in the numerator: $L = \mathop {\lim }\limits_{x \to {\pi \over 2}} {{\cos x \left( \frac{1}{\sin x} - 1 \right)} \over {{{\left( {\pi - 2x} \right)}^3}}}$ $L = \mathop {\lim }\limits_{x \to {\pi \over 2}} {{\cos x \left( \frac{1 - \sin x}{\sin x} \right)} \over {{{\left( {\pi - 2x} \right)}^3}}}$ $L = \mathop {\lim }\limits_{x \to {\pi \over 2}} {{\cos x (1 - \sin x)} \over {\sin x {{\left( {\pi - 2x} \right)}^3}}}$ As $x \to \frac{\pi}{2}$, we have: $\cos x \to \cos(\frac{\pi}{2}) = 0$ $\sin x \to \sin(\frac{\pi}{2}) = 1$ $1 - \sin x \to 1 - 1 = 0$ $\pi - 2x \to \pi - 2(\frac{\pi}{2}) = \pi - \pi = 0$ So, the numerator approaches $0 \times (1-1) = 0$, and the denominator approaches $1 \times 0^3 = 0$. This is an indeterminate form of type $\frac{0}{0}$.

Step 2: Introduce a substitution to simplify the denominator. The term $(\pi - 2x)^3$ in the denominator suggests a substitution related to $\frac{\pi}{2} - x$. Let $t = \frac{\pi}{2} - x$. As $x \to \frac{\pi}{2}$, we have $t \to \frac{\pi}{2} - \frac{\pi}{2} = 0$. From $t = \frac{\pi}{2} - x$, we can express $x$ as $x = \frac{\pi}{2} - t$. Also, $\pi - 2x = \pi - 2(\frac{\pi}{2} - t) = \pi - \pi + 2t = 2t$. Therefore, $(\pi - 2x)^3 = (2t)^3 = 8t^3$.

Now, we need to express the trigonometric terms in the numerator in terms of $t$: $\cos x = \cos(\frac{\pi}{2} - t) = \sin t$ $\sin x = \sin(\frac{\pi}{2} - t) = \cos t$

Substitute these into the limit expression: $L = \mathop {\lim }\limits_{t \to 0} {{\sin t (1 - \cos t)} \over {\cos t (2t)^3}}$ $L = \mathop {\lim }\limits_{t \to 0} {{\sin t (1 - \cos t)} \over {8t^3 \cos t}}$ We can pull out the constant factor $\frac{1}{8}$: $L = \frac{1}{8} \mathop {\lim }\limits_{t \to 0} {{\sin t (1 - \cos t)} \over {t^3 \cos t}}$

Step 3: Use trigonometric identities to simplify further and apply standard limits. We know the identity $1 - \cos t = 2 \sin^2(\frac{t}{2})$. Substitute this into the expression: $L = \frac{1}{8} \mathop {\lim }\limits_{t \to 0} {{\sin t (2 \sin^2(\frac{t}{2}))} \over {t^3 \cos t}}$ $L = \frac{1}{8} \mathop {\lim }\limits_{t \to 0} {{2 \sin t \sin^2(\frac{t}{2})} \over {t^3 \cos t}}$ Pull out the constant factor $2$ from the numerator: $L = \frac{2}{8} \mathop {\lim }\limits_{t \to 0} {{\sin t \sin^2(\frac{t}{2})} \over {t^3 \cos t}}$ $L = \frac{1}{4} \mathop {\lim }\limits_{t \to 0} {{\sin t \sin^2(\frac{t}{2})} \over {t^3 \cos t}}$ To apply the standard limit $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$, we need to rearrange the terms: $L = \frac{1}{4} \mathop {\lim }\limits_{t \to 0} \left( \frac{\sin t}{t} \cdot \frac{\sin^2(\frac{t}{2})}{t^2} \cdot \frac{1}{\cos t} \right)$ We need to adjust the denominator of the $\sin^2(\frac{t}{2})$ term. The current denominator is $t^2$. We want it to be $(\frac{t}{2})^2 = \frac{t^2}{4}$. So, we can rewrite $\frac{\sin^2(\frac{t}{2})}{t^2}$ as $\frac{\sin^2(\frac{t}{2})}{(\frac{t}{2})^2 \cdot 4} = \frac{1}{4} \left(\frac{\sin(\frac{t}{2})}{\frac{t}{2}}\right)^2$.

Let's rewrite the expression carefully: $L = \frac{1}{4} \mathop {\lim }\limits_{t \to 0} \left( \frac{\sin t}{t} \cdot \frac{\sin^2(\frac{t}{2})}{t^2} \cdot \frac{1}{\cos t} \right)$ $L = \frac{1}{4} \mathop {\lim }\limits_{t \to 0} \left( \frac{\sin t}{t} \cdot \frac{\sin^2(\frac{t}{2})}{(\frac{t}{2})^2 \cdot 4} \cdot \frac{1}{\cos t} \right)$ $L = \frac{1}{4} \mathop {\lim }\limits_{t \to 0} \left( \frac{\sin t}{t} \cdot \frac{1}{4} \left(\frac{\sin(\frac{t}{2})}{\frac{t}{2}}\right)^2 \cdot \frac{1}{\cos t} \right)$ Pull out the constant factor $\frac{1}{4}$: $L = \frac{1}{4} \cdot \frac{1}{4} \mathop {\lim }\limits_{t \to 0} \left( \frac{\sin t}{t} \cdot \left(\frac{\sin(\frac{t}{2})}{\frac{t}{2}}\right)^2 \cdot \frac{1}{\cos t} \right)$ $L = \frac{1}{16} \mathop {\lim }\limits_{t \to 0} \left( \frac{\sin t}{t} \cdot \left(\frac{\sin(\frac{t}{2})}{\frac{t}{2}}\right)^2 \cdot \frac{1}{\cos t} \right)$

Step 4: Evaluate the limit by substituting the values of the standard limits. Now, we can evaluate each part of the product as $t \to 0$:

• $\mathop {\lim }\limits_{t \to 0} \frac{\sin t}{t} = 1$
• As $t \to 0$, $\frac{t}{2} \to 0$. So, $\mathop {\lim }\limits_{t \to 0} \frac{\sin(\frac{t}{2})}{\frac{t}{2}} = 1$. Therefore, $\mathop {\lim }\limits_{t \to 0} \left(\frac{\sin(\frac{t}{2})}{\frac{t}{2}}\right)^2 = 1^2 = 1$.
• $\mathop {\lim }\limits_{t \to 0} \cos t = \cos(0) = 1$. So, $\mathop {\lim }\limits_{t \to 0} \frac{1}{\cos t} = \frac{1}{1} = 1$.

Substitute these values back into the expression for $L$: $L = \frac{1}{16} \times (1) \times (1)^2 \times (1)$ $L = \frac{1}{16}$

Common Mistakes & Tips

• Algebraic Errors: Be very careful with algebraic manipulations, especially when combining fractions or factoring. A small mistake can lead to an incorrect final answer.
• Incorrect Substitution: Ensure that when you make a substitution (like $t = \frac{\pi}{2} - x$), you correctly transform all parts of the expression (numerator, denominator, and the limit point).
• Misapplication of Standard Limits: Make sure the arguments of the sine functions match their denominators in the standard limit $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. For example, $\sin(\frac{t}{2})$ needs to be divided by $\frac{t}{2}$, not just $t$.
• L'Hôpital's Rule Caution: While L'Hôpital's rule can be used, it might involve differentiating complex functions multiple times, increasing the chances of errors. The method using standard limits is often more elegant and safer for this type of problem.

Summary

The problem involves evaluating a limit that results in an indeterminate form $\frac{0}{0}$. We began by rewriting the trigonometric functions and simplifying the numerator. A crucial step was to introduce a substitution $t = \frac{\pi}{2} - x$ to transform the limit to $t \to 0$, which simplifies the denominator. We then utilized trigonometric identities, specifically $1 - \cos t = 2 \sin^2(\frac{t}{2})$, to break down the expression into forms that match standard limits like $\mathop {\lim }\limits_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$. By carefully rearranging terms and applying these standard limits, we arrived at the final value of the limit.

The final answer is $\boxed{{1 \over {16}}}$.