Question

$\lim_{x \to \frac{\pi}{4}} \frac{\cot^3x-\tan x}{\cos(x+\frac{\pi}{4}) \cos(x+\frac{\pi}{4})}$ is:

• A. $4\sqrt{2}$
• B. 8
• C. 4
• D. $8\sqrt{2}$

Answer 1

Answer: (B)

8

Answer 2

To evaluate the limit $\lim_{x \to \frac{\pi}{4}} \frac{\cot^3x-\tan x}{\cos^2(x+\frac{\pi}{4})}$, we first check the form of the limit by substituting $x = \frac{\pi}{4}$:

Numerator: $\cot^3(\frac{\pi}{4}) - \tan(\frac{\pi}{4}) = 1^3 - 1 = 0$.

Denominator: $\cos^2(\frac{\pi}{4}+\frac{\pi}{4}) = \cos^2(\frac{\pi}{2}) = 0^2 = 0$.

The limit is of the indeterminate form $\frac{0}{0}$. We can use L'Hopital's Rule or algebraic manipulation.

Method 1: Using L'Hopital's Rule

Let $f(x) = \cot^3x-\tan x$ and $g(x) = \cos^2(x+\frac{\pi}{4})$.

First derivatives:

$f'(x) = \frac{d}{dx}(\cot^3x-\tan x) = 3\cot^2x(-\csc^2x) - \sec^2x = -3\cot^2x\csc^2x - \sec^2x$.

$g'(x) = \frac{d}{dx}(\cos^2(x+\frac{\pi}{4})) = 2\cos(x+\frac{\pi}{4})(-\sin(x+\frac{\pi}{4})) = -2\sin(x+\frac{\pi}{4})\cos(x+\frac{\pi}{4})$.

Using the identity $\sin(2A) = 2\sin A \cos A$, we have $g'(x) = -\sin(2(x+\frac{\pi}{4})) = -\sin(2x+\frac{\pi}{2})$.

Using the identity $\sin(\theta+\frac{\pi}{2}) = \cos\theta$, we get $g'(x) = -\cos(2x)$.

Evaluate $f'(\frac{\pi}{4})$ and $g'(\frac{\pi}{4})$:

$f'(\frac{\pi}{4}) = -3\cot^2(\frac{\pi}{4})\csc^2(\frac{\pi}{4}) - \sec^2(\frac{\pi}{4}) = -3(1)^2(\sqrt{2})^2 - (\sqrt{2})^2 = -3(2) - 2 = -6 - 2 = -8$.

$g'(\frac{\pi}{4}) = -\cos(2 \cdot \frac{\pi}{4}) = -\cos(\frac{\pi}{2}) = 0$.

Since $g'(\frac{\pi}{4})=0$, the limit is still of the form $\frac{-8}{0}$, which means we need to apply L'Hopital's Rule again.

Second derivatives:

$f''(x) = \frac{d}{dx}(-3\cot^2x\csc^2x - \sec^2x)$ $= -3[2\cot x(-\csc^2x)\csc^2x + \cot^2x(2\csc x)(-\csc x\cot x)] - 2\sec x(\sec x\tan x)$ $= -3[-2\cot x\csc^4x - 2\cot^3x\csc^2x] - 2\sec^2x\tan x$ $= 6\cot x\csc^4x + 6\cot^3x\csc^2x - 2\sec^2x\tan x$.

Evaluate $f''(\frac{\pi}{4})$:

$f''(\frac{\pi}{4}) = 6(1)(\sqrt{2})^4 + 6(1)^3(\sqrt{2})^2 - 2(\sqrt{2})^2(1)$ $= 6(4) + 6(2) - 2(2) = 24 + 12 - 4 = 32$.

$g''(x) = \frac{d}{dx}(-\cos(2x)) = -(-\sin(2x) \cdot 2) = 2\sin(2x)$.

Evaluate $g''(\frac{\pi}{4})$:

$g''(\frac{\pi}{4}) = 2\sin(2 \cdot \frac{\pi}{4}) = 2\sin(\frac{\pi}{2}) = 2(1) = 2$.

Now, apply L'Hopital's Rule:

$\lim_{x \to \frac{\pi}{4}} \frac{f''(x)}{g''(x)} = \frac{32}{2} = 16$.

However, 16 is not among the options. Let's recheck the problem statement or try algebraic manipulation more carefully.

Method 2: Using Algebraic Manipulation

Let $x = \frac{\pi}{4} + h$. As $x \to \frac{\pi}{4}$, $h \to 0$.

The limit becomes $\lim_{h \to 0} \frac{\cot^3(\frac{\pi}{4}+h)-\tan(\frac{\pi}{4}+h)}{\cos^2(\frac{\pi}{4}+h+\frac{\pi}{4})}$.

Denominator: $\cos^2(\frac{\pi}{2}+h) = (-\sin h)^2 = \sin^2 h$.

Numerator:

We know $\tan(\frac{\pi}{4}+h) = \frac{1+\tan h}{1-\tan h}$ and $\cot(\frac{\pi}{4}+h) = \frac{1-\tan h}{1+\tan h}$.

Let $t = \tan h$. As $h \to 0$, $t \to 0$.

Numerator is $\left(\frac{1-t}{1+t}\right)^3 - \frac{1+t}{1-t} = \frac{(1-t)^4 - (1+t)^4}{(1+t)^3(1-t)}$.

Using the identity $a^4-b^4 = (a^2-b^2)(a^2+b^2) = (a-b)(a+b)(a^2+b^2)$:

$(1-t)^4 - (1+t)^4 = ((1-t)^2 - (1+t)^2)((1-t)^2 + (1+t)^2)$ $= ((1-2t+t^2) - (1+2t+t^2))((1-2t+t^2) + (1+2t+t^2))$ $= (-4t)(2+2t^2) = -8t(1+t^2)$.

So, the numerator is $\frac{-8t(1+t^2)}{(1+t)^3(1-t)}$.

The expression becomes $\lim_{t \to 0} \frac{\frac{-8t(1+t^2)}{(1+t)^3(1-t)}}{\sin^2 h}$.

Since $t = \tan h$, $\sin^2 h = \left(\frac{\tan h}{\sqrt{1+\tan^2 h}}\right)^2 = \frac{\tan^2 h}{1+\tan^2 h} = \frac{t^2}{1+t^2}$.

So the limit is $\lim_{t \to 0} \frac{-8t(1+t^2)}{(1+t)^3(1-t)} \cdot \frac{1+t^2}{t^2}$. $= \lim_{t \to 0} \frac{-8(1+t^2)^2}{t(1+t)^3(1-t)}$.

This still results in $\frac{-8}{0}$, which means the limit is $\infty$ or $-\infty$.

Let's re-examine the denominator of the original expression: $\cos(x+\frac{\pi}{4}) \cos(x+\frac{\pi}{4})$.

The question writes it as $\cos(x+\frac{\pi}{4}) \cos(x+\frac{\pi}{4})$, which is $\cos^2(x+\frac{\pi}{4})$. This is what I used.

Let's re-read the question carefully. Is there a typo?

If the denominator was $\cos(x+\frac{\pi}{4})$, then $g(x) = \cos(x+\frac{\pi}{4})$.

$g'(x) = -\sin(x+\frac{\pi}{4})$.

$g'(\frac{\pi}{4}) = -\sin(\frac{\pi}{2}) = -1$.

In this case, the limit would be $\frac{f'(\frac{\pi}{4})}{g'(\frac{\pi}{4})} = \frac{-8}{-1} = 8$.

This is option (2).

It is common for questions to have a typo where a square is intended but written as a product of two identical terms. If it were $\cos^2(x+\frac{\pi}{4})$, the answer would be 16. If it were $\cos(x+\frac{\pi}{4})$, the answer would be 8. Given the options, it is highly likely that the denominator was intended to be $\cos(x+\frac{\pi}{4})$.

Let's assume the denominator is $\cos(x+\frac{\pi}{4})$.

$\lim_{x \to \frac{\pi}{4}} \frac{\cot^3x-\tan x}{\cos(x+\frac{\pi}{4})}$

Numerator $f(x) = \cot^3x-\tan x$. $f(\frac{\pi}{4})=0$.

$f'(x) = -3\cot^2x\csc^2x - \sec^2x$.

$f'(\frac{\pi}{4}) = -3(1)^2(\sqrt{2})^2 - (\sqrt{2})^2 = -3(2) - 2 = -6 - 2 = -8$.

Denominator $g(x) = \cos(x+\frac{\pi}{4})$. $g(\frac{\pi}{4})=0$.

$g'(x) = -\sin(x+\frac{\pi}{4})$.

$g'(\frac{\pi}{4}) = -\sin(\frac{\pi}{4}+\frac{\pi}{4}) = -\sin(\frac{\pi}{2}) = -1$.

Using L'Hopital's Rule:

$\lim_{x \to \frac{\pi}{4}} \frac{f'(x)}{g'(x)} = \frac{-8}{-1} = 8$.

This matches option (2). This is a common situation in multiple choice questions where a slight ambiguity in notation or a typo can lead to one of the given options. The repeated term $\cos(x+\frac{\pi}{4}) \cos(x+\frac{\pi}{4})$ strongly suggests $\cos^2(x+\frac{\pi}{4})$, but if that leads to an answer not in options, then it's worth considering the possibility of a simpler denominator.

Let's verify using the algebraic method for the simpler denominator: $\lim_{h \to 0} \frac{\frac{-8t(1+t^2)}{(1+t)^3(1-t)}}{\sin h}$. Since $t = \tan h$, $\sin h \approx h$ and $t \approx h$ for small $h$. $\lim_{t \to 0} \frac{-8t(1+t^2)}{(1+t)^3(1-t) \cdot t}$. Cancel $t$: $\lim_{t \to 0} \frac{-8(1+t^2)}{(1+t)^3(1-t)}$. Substitute $t=0$: $\frac{-8(1+0)^2}{(1+0)^3(1-0)} = \frac{-8}{1 \cdot 1} = -8$.

Wait, there is a sign error in the algebraic method. $\cos(x+\frac{\pi}{4}) = \cos(\frac{\pi}{2}+h) = -\sin h$. So the denominator is $-\sin h$. The expression becomes $\lim_{h \to 0} \frac{\frac{-8t(1+t^2)}{(1+t)^3(1-t)}}{-\sin h}$. $= \lim_{t \to 0} \frac{-8t(1+t^2)}{(1+t)^3(1-t)} \cdot \frac{1}{-\sin h}$. Since $t=\tan h$, $\sin h = \frac{\tan h}{\sqrt{1+\tan^2 h}} = \frac{t}{\sqrt{1+t^2}}$. So, $\frac{1}{-\sin h} = \frac{-\sqrt{1+t^2}}{t}$. The limit is $\lim_{t \to 0} \frac{-8t(1+t^2)}{(1+t)^3(1-t)} \cdot \frac{-\sqrt{1+t^2}}{t}$. Cancel $t$: $= \lim_{t \to 0} \frac{-8(1+t^2)(- \sqrt{1+t^2})}{(1+t)^3(1-t)}$. $= \lim_{t \to 0} \frac{8(1+t^2)^{3/2}}{(1+t)^3(1-t)}$. Substitute $t=0$: $\frac{8(1+0)^{3/2}}{(1+0)^3(1-0)} = \frac{8(1)}{1 \cdot 1} = 8$.

Both methods (L'Hopital's and algebraic manipulation) yield 8, assuming the denominator is $\cos(x+\frac{\pi}{4})$. This strongly suggests that the question intended the denominator to be $\cos(x+\frac{\pi}{4})$ rather than $\cos^2(x+\frac{\pi}{4})$.