Key Concepts and Formulas

• Limit Evaluation: When a limit results in an indeterminate form (like 0/0 or $\infty/\infty$), we can use algebraic manipulation, trigonometric identities, or L'Hôpital's Rule to evaluate it.
• Trigonometric Identities:
  • $\cot x = \frac{1}{\tan x}$
  • $\tan^2 x + 1 = \sec^2 x$
  • $\cos(A+B) = \cos A \cos B - \sin A \sin B$
  • $\cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$
• Special Trigonometric Values:
  • $\tan(\frac{\pi}{4}) = 1$
  • $\cot(\frac{\pi}{4}) = 1$
  • $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$
  • $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$
  • $\sec(\frac{\pi}{4}) = \sqrt{2}$
  • $\csc(\frac{\pi}{4}) = \sqrt{2}$
• L'Hôpital's Rule: If $\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)}$ is of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$, then $\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)} = \mathop {\lim }\limits_{x \to c} \frac{f'(x)}{g'(x)}$, provided the latter limit exists.

Step-by-Step Solution

We want to evaluate the limit: $L = \mathop {\lim }\limits_{x \to \pi /4} {{{{\cot }^3}x - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$

Step 1: Check for Indeterminate Form Substitute $x = \frac{\pi}{4}$ into the expression: Numerator: $(\cot(\frac{\pi}{4}))^3 - \tan(\frac{\pi}{4}) = 1^3 - 1 = 1 - 1 = 0$ Denominator: $\cos(\frac{\pi}{4} + \frac{\pi}{4}) = \cos(\frac{\pi}{2}) = 0$ Since we have the indeterminate form $\frac{0}{0}$, we can proceed with algebraic manipulation or L'Hôpital's Rule.

Step 2: Algebraic Manipulation - Express in terms of $\tan x$ Rewrite $\cot x$ as $\frac{1}{\tan x}$ to work with a single trigonometric function in the numerator. $L = \mathop {\lim }\limits_{x \to \pi /4} {{{1 \over {{{\tan }^3}x}} - \tan x} \over {\cos \left( {x + {\pi \over 4}} \right)}}$

Step 3: Combine terms in the numerator Find a common denominator for the terms in the numerator. $L = \mathop {\lim }\limits_{x \to \pi /4} {{{1 - {{\tan }^4}x} \over {{{\tan }^3}x}} \over {\cos \left( {x + {\pi \over 4}} \right)}}$ $L = \mathop {\lim }\limits_{x \to \pi /4} {{{1 - {{\tan }^4}x} \over {{{\tan }^3}x \cos \left( {x + {\pi \over 4}} \right)}}}$

Step 4: Factor the numerator Use the difference of squares formula, $a^2 - b^2 = (a-b)(a+b)$, where $a=1$ and $b=\tan^2 x$. Then use $1+\tan^2 x = \sec^2 x$. $L = \mathop {\lim }\limits_{x \to \pi /4} {{\left( {1 - {{\tan }^2}x} \right)\left( {1 + {{\tan }^2}x} \right)} \over {{{\tan }^3}x \cos \left( {x + {\pi \over 4}} \right)}}$

Step 5: Expand the denominator $\cos(x + \frac{\pi}{4})$ Use the angle addition formula for cosine: $\cos(A+B) = \cos A \cos B - \sin A \sin B$. Here, $A=x$ and $B=\frac{\pi}{4}$. $\cos \left( {x + {\pi \over 4}} \right) = \cos x \cos \frac{\pi}{4} - \sin x \sin \frac{\pi}{4}$ $= \cos x \left(\frac{1}{\sqrt{2}}\right) - \sin x \left(\frac{1}{\sqrt{2}}\right)$ $= \frac{1}{\sqrt{2}} (\cos x - \sin x)$

Step 6: Substitute the expanded denominator and simplify Substitute this back into the limit expression. $L = \mathop {\lim }\limits_{x \to \pi /4} {{{{\left( {1 - {{\tan }^2}x} \right)\left( {1 + {{\tan }^2}x} \right)}} \over {{{\tan }^3}x \left( \frac{1}{\sqrt{2}} (\cos x - \sin x) \right)}}}$ $L = \mathop {\lim }\limits_{x \to \pi /4} {{{\sqrt{2}} \left( {1 - {{\tan }^2}x} \right)\left( {1 + {{\tan }^2}x} \right)} \over {{{\tan }^3}x (\cos x - \sin x)}}$

Step 7: Express $\tan^2 x$ in terms of $\sin x$ and $\cos x$ Recall that $\tan x = \frac{\sin x}{\cos x}$, so $\tan^2 x = \frac{\sin^2 x}{\cos^2 x}$. $1 - \tan^2 x = 1 - \frac{\sin^2 x}{\cos^2 x} = \frac{\cos^2 x - \sin^2 x}{\cos^2 x}$ And $1 + \tan^2 x = \sec^2 x = \frac{1}{\cos^2 x}$.

Step 8: Substitute these into the limit expression $L = \mathop {\lim }\limits_{x \to \pi /4} {{{\sqrt{2}} \left( \frac{\cos^2 x - \sin^2 x}{\cos^2 x} \right) \left( \frac{1}{\cos^2 x} \right)} \over {{{\left(\frac{\sin x}{\cos x}\right) }^3} (\cos x - \sin x)}}$ $L = \mathop {\lim }\limits_{x \to \pi /4} {{{\sqrt{2}} \left( \cos^2 x - \sin^2 x \right)} \over {{{\cos }^6}x}} \times \frac{{{{\cos }^3}x}}{{{{\sin }^3}x (\cos x - \sin x)}}$ $L = \mathop {\lim }\limits_{x \to \pi /4} {{{\sqrt{2}} \left( \cos^2 x - \sin^2 x \right)} \over {{{\cos }^3}x \sin^3 x (\cos x - \sin x)}}$

Step 9: Factor the difference of squares in the numerator Use $\cos^2 x - \sin^2 x = (\cos x - \sin x)(\cos x + \sin x)$. $L = \mathop {\lim }\limits_{x \to \pi /4} {{{\sqrt{2}} (\cos x - \sin x)(\cos x + \sin x)} \over {{{\cos }^3}x \sin^3 x (\cos x - \sin x)}}$

Step 10: Cancel the common term $(\cos x - \sin x)$ Since $x \to \frac{\pi}{4}$, $x \neq \frac{\pi}{4}$, so $\cos x - \sin x \neq 0$. $L = \mathop {\lim }\limits_{x \to \pi /4} {{{\sqrt{2}} (\cos x + \sin x)} \over {{{\cos }^3}x \sin^3 x}}$

Step 11: Evaluate the limit by substituting $x = \frac{\pi}{4}$ Now, substitute the values of trigonometric functions at $x = \frac{\pi}{4}$. $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ $L = \frac{{\sqrt{2}} \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right)}{{{\left(\frac{1}{\sqrt{2}}\right) }^3} {\left(\frac{1}{\sqrt{2}}\right) }^3}$ $L = \frac{{\sqrt{2}} \left(\frac{2}{\sqrt{2}}\right)}{{\frac{1}{2\sqrt{2}}} {\frac{1}{2\sqrt{2}}}}$ $L = \frac{2}{{\frac{1}{8}}}$ $L = 2 \times 8$ $L = 16$

Let me recheck the steps. There seems to be a calculation error.

Let's re-evaluate from Step 10.

Step 10 (Revised): Cancel the common term $(\cos x - \sin x)$ Since $x \to \frac{\pi}{4}$, $x \neq \frac{\pi}{4}$, so $\cos x - \sin x \neq 0$. $L = \mathop {\lim }\limits_{x \to \pi /4} {{{\sqrt{2}} (\cos x + \sin x)} \over {{{\cos }^3}x \sin^3 x}}$

Step 11 (Revised): Evaluate the limit by substituting $x = \frac{\pi}{4}$ Substitute the values of trigonometric functions at $x = \frac{\pi}{4}$. $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ $L = \frac{{\sqrt{2}} \left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right)}{{{\left(\frac{1}{\sqrt{2}}\right) }^3} {\left(\frac{1}{\sqrt{2}}\right) }^3}$ $L = \frac{{\sqrt{2}} \left(\frac{2}{\sqrt{2}}\right)}{{\left(\frac{1}{2\sqrt{2}}\right)} \times {\left(\frac{1}{2\sqrt{2}}\right)}}$ $L = \frac{2}{{\frac{1}{8}}}$ $L = 16$

There is still a discrepancy with the provided answer. Let's review the algebraic manipulation.

Let's try L'Hôpital's Rule as an alternative approach to verify.

Alternative Method: L'Hôpital's Rule

Step A1: Verify Indeterminate Form As checked in Step 1, the limit is of the form $\frac{0}{0}$.

Step A2: Differentiate Numerator and Denominator Let $f(x) = \cot^3 x - \tan x$ and $g(x) = \cos(x + \frac{\pi}{4})$. $f'(x) = \frac{d}{dx}(\cot^3 x - \tan x) = 3\cot^2 x (-\csc^2 x) - \sec^2 x$ $g'(x) = \frac{d}{dx}(\cos(x + \frac{\pi}{4})) = -\sin(x + \frac{\pi}{4}) \cdot 1 = -\sin(x + \frac{\pi}{4})$

Step A3: Apply L'Hôpital's Rule $L = \mathop {\lim }\limits_{x \to \pi /4} \frac{f'(x)}{g'(x)} = \mathop {\lim }\limits_{x \to \pi /4} \frac{3\cot^2 x (-\csc^2 x) - \sec^2 x}{-\sin(x + \frac{\pi}{4})}$

Step A4: Substitute $x = \frac{\pi}{4}$ Recall the values: $\cot(\frac{\pi}{4}) = 1$ $\csc(\frac{\pi}{4}) = \sqrt{2}$ $\sec(\frac{\pi}{4}) = \sqrt{2}$ $\sin(\frac{\pi}{4} + \frac{\pi}{4}) = \sin(\frac{\pi}{2}) = 1$

$L = \frac{3(1)^2 (-(\sqrt{2})^2) - (\sqrt{2})^2}{-(1)}$ $L = \frac{3(1)(-2) - 2}{-1}$ $L = \frac{-6 - 2}{-1}$ $L = \frac{-8}{-1}$ $L = 8$

This matches the correct answer. Let's review the algebraic steps for the error.

Reviewing Algebraic Steps

Let's go back to: $L = \mathop {\lim }\limits_{x \to \pi /4} {{{1 - {{\tan }^4}x} \over {{{\tan }^3}x \cos \left( {x + {\pi \over 4}} \right)}}}$

And the expansion of the denominator: $\cos \left( {x + {\pi \over 4}} \right) = \frac{1}{\sqrt{2}} (\cos x - \sin x)$

So, $L = \mathop {\lim }\limits_{x \to \pi /4} {{{1 - {{\tan }^4}x} \over {{{\tan }^3}x \frac{1}{\sqrt{2}} (\cos x - \sin x)}}}$ $L = \mathop {\lim }\limits_{x \to \pi /4} {{{\sqrt{2}} (1 - {{\tan }^4}x)} \over {{{\tan }^3}x (\cos x - \sin x)}}$

We used $1 - \tan^4 x = (1-\tan^2 x)(1+\tan^2 x)$. Also, $1+\tan^2 x = \sec^2 x$. So, $1 - \tan^4 x = (1-\tan^2 x)\sec^2 x$.

And $\tan^3 x = \frac{\sin^3 x}{\cos^3 x}$.

$L = \mathop {\lim }\limits_{x \to \pi /4} {{{\sqrt{2}} (1 - {{\tan }^2}x)\sec^2 x} \over {{{\tan }^3}x (\cos x - \sin x)}}$

Substitute $\tan x = \frac{\sin x}{\cos x}$ and $\sec x = \frac{1}{\cos x}$. $1 - \tan^2 x = 1 - \frac{\sin^2 x}{\cos^2 x} = \frac{\cos^2 x - \sin^2 x}{\cos^2 x}$ $\sec^2 x = \frac{1}{\cos^2 x}$ $\tan^3 x = \frac{\sin^3 x}{\cos^3 x}$

$L = \mathop {\lim }\limits_{x \to \pi /4} {{{\sqrt{2}} \left( \frac{\cos^2 x - \sin^2 x}{\cos^2 x} \right) \left( \frac{1}{\cos^2 x} \right)} \over {\left( \frac{\sin^3 x}{\cos^3 x} \right) (\cos x - \sin x)}}$

$L = \mathop {\lim }\limits_{x \to \pi /4} {{{\sqrt{2}} (\cos^2 x - \sin^2 x)} \over {{\cos }^4 x}} \times \frac{{\cos }^3 x \sin^3 x}{\sin^3 x (\cos x - \sin x)}$

This step seems to be where the error occurred in the original provided solution. The $\cos^3 x$ from the denominator of $\tan^3 x$ was incorrectly handled.

Let's correct from the expression: $L = \mathop {\lim }\limits_{x \to \pi /4} {{{\sqrt{2}} (1 - {{\tan }^4}x)} \over {{{\tan }^3}x (\cos x - \sin x)}}$ $L = \mathop {\lim }\limits_{x \to \pi /4} {{{\sqrt{2}} (1 - {{\tan }^2}x)(1 + {{\tan }^2}x)} \over {{{\tan }^3}x (\cos x - \sin x)}}$

Use $1-\tan^2 x = \frac{\cos^2 x - \sin^2 x}{\cos^2 x}$ and $1+\tan^2 x = \sec^2 x$. $L = \mathop {\lim }\limits_{x \to \pi /4} {{{\sqrt{2}} \left( \frac{\cos^2 x - \sin^2 x}{\cos^2 x} \right) \left( \sec^2 x \right)} \over {{{\tan }^3}x (\cos x - \sin x)}}$ $L = \mathop {\lim }\limits_{x \to \pi /4} {{{\sqrt{2}} \left( \frac{(\cos x - \sin x)(\cos x + \sin x)}{\cos^2 x} \right) \left( \frac{1}{\cos^2 x} \right)} \over {{{\tan }^3}x (\cos x - \sin x)}}$

Cancel $(\cos x - \sin x)$: $L = \mathop {\lim }\limits_{x \to \pi /4} {{{\sqrt{2}} (\cos x + \sin x)} \over {{\cos }^4 x \tan^3 x (\cos x - \sin x)}}$ There is still a mistake. Let's restart the algebraic manipulation from a cleaner point.

Step 2 (Revised): Express in terms of $\sin x$ and $\cos x$ $L = \mathop {\lim }\limits_{x \to \pi /4} {{{{{\left( {\frac{\cos x}{\sin x}} \right)}}^3} - \frac{\sin x}{\cos x}} \over {\cos \left( {x + {\pi \over 4}} \right)}}$ L = \mathop {\lim }\limits_{x \to \pi /4} {{{\frac{{\cos }^3 x}}{{{\sin }^3 x}} - \frac{\sin x}{\cos x}} \over {\cos \left( {x + {\pi \over 4}} \right)}}

Step 3 (Revised): Combine terms in the numerator $L = \mathop {\lim }\limits_{x \to \pi /4} {{{\frac{{\cos }^4 x - {{\sin }^4}x}{{{\sin }^3}x {\cos x}}} \over {\cos \left( {x + {\pi \over 4}} \right)}}}$ $L = \mathop {\lim }\limits_{x \to \pi /4} {{{\cos }^4 x - {{\sin }^4}x} \over {{{\sin }^3}x {\cos x} \cos \left( {x + {\pi \over 4}} \right)}}$

Step 4 (Revised): Factor the numerator Use the difference of squares: $\cos^4 x - \sin^4 x = (\cos^2 x - \sin^2 x)(\cos^2 x + \sin^2 x) = (\cos^2 x - \sin^2 x)(1)$. $L = \mathop {\lim }\limits_{x \to \pi /4} {{{\cos^2 x - \sin^2 x}} \over {{{\sin }^3}x {\cos x} \cos \left( {x + {\pi \over 4}} \right)}}$

Step 5 (Revised): Expand the denominator $\cos(x + \frac{\pi}{4})$ $\cos \left( {x + {\pi \over 4}} \right) = \frac{1}{\sqrt{2}} (\cos x - \sin x)$

Step 6 (Revised): Substitute and simplify $L = \mathop {\lim }\limits_{x \to \pi /4} {{{\cos^2 x - \sin^2 x}} \over {{{\sin }^3}x {\cos x} \left( \frac{1}{\sqrt{2}} (\cos x - \sin x) \right)}}$ L = \mathop {\lim }\limits_{x \to \pi /4} {{{\sqrt{2}}(\cos^2 x - \sin^2 x)}} \over {{{\sin }^3}x {\cos x} (\cos x - \sin x)}}

Step 7 (Revised): Factor $\cos^2 x - \sin^2 x$ $L = \mathop {\lim }\limits_{x \to \pi /4} {{{\sqrt{2}}(\cos x - \sin x)(\cos x + \sin x)}} \over {{{\sin }^3}x {\cos x} (\cos x - \sin x)}$

Step 8 (Revised): Cancel $(\cos x - \sin x)$ L = \mathop {\lim }\limits_{x \to \pi /4} {{{\sqrt{2}}(\cos x + \sin x)}} \over {{{\sin }^3}x {\cos x}}}

Step 9 (Revised): Evaluate the limit Substitute $x = \frac{\pi}{4}$: $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ $\sin(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$ $L = \frac{{\sqrt{2}}\left(\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}\right)}{{{\left(\frac{1}{\sqrt{2}}\right) }^3} {\left(\frac{1}{\sqrt{2}}\right)}}$ $L = \frac{{\sqrt{2}}\left(\frac{2}{\sqrt{2}}\right)}{{\frac{1}{2\sqrt{2}}} \times \frac{1}{\sqrt{2}}}$ $L = \frac{2}{{\frac{1}{4}}}$ $L = 2 \times 4$ $L = 8$

This algebraic method now correctly yields 8. The error in the original solution was in the simplification of terms involving $\tan^3 x$ and $\cos^2 x$ in the denominator.

Common Mistakes & Tips

• Algebraic Errors: Be extremely careful when manipulating trigonometric expressions, especially when combining fractions and canceling terms. A small mistake can lead to a significantly different result.
• L'Hôpital's Rule Applicability: Ensure the limit is in an indeterminate form ($\frac{0}{0}$ or $\frac{\infty}{\infty}$) before applying L'Hôpital's Rule. Differentiating a determinate form will give an incorrect result.
• Trigonometric Identities: Have key trigonometric identities memorized, as they are crucial for simplifying complex expressions in limits.
• Special Angle Values: Knowing the values of trigonometric functions for common angles like $\frac{\pi}{4}$ is essential for evaluating limits.

Summary

The limit was evaluated by first checking for an indeterminate form ($\frac{0}{0}$). We then used algebraic manipulation by expressing all trigonometric functions in terms of sine and cosine. By factoring the numerator and expanding the denominator, we were able to cancel out the term causing the indeterminate form. Finally, direct substitution of $x = \frac{\pi}{4}$ into the simplified expression yielded the result. Alternatively, L'Hôpital's Rule was applied by differentiating the numerator and denominator separately, which also led to the correct answer.

The final answer is $\boxed{8}$. This corresponds to option (D).