The value of `lim_(xto4)((cos alpha)^(x)-(sinalpha)^(x)-cos 2alpha)/(x-4), alpha epsilon(0,(pi)/2)` is

Options

A. `log(cos alpha)+(sin alpha)^(4)log(sin alpha)`
B. `(cos^(4)alpha)log(cos alpha)-(sin alpha)^(4)log(sinalpha)`
C. `(cos^(4)alpha)log(cos alpha)`
D. None of these

Answer

To solve the limit problem, we need to evaluate the limit:

\[ \lim_{x \to 4} \frac{(\cos \alpha)^x - (\sin \alpha)^x - \cos 2\alpha}{x - 4} \]

**Step 1: Check the form of the limit**

First, we substitute \( x = 4 \) into the numerator:

\[ (\cos \alpha)^4 - (\sin \alpha)^4 - \cos 2\alpha \]

We need to check if this results in \( 0 \):

\[ \cos^4 \alpha - \sin^4 \alpha - \cos 2\alpha \]

Using the identity \( \cos^4 \alpha - \sin^4 \alpha = (\cos^2 \alpha + \sin^2 \alpha)(\cos^2 \alpha - \sin^2 \alpha) \) and knowing that \( \cos^2 \alpha + \sin^2 \alpha = 1 \):

\[ \cos^4 \alpha - \sin^4 \alpha = 1 \cdot (\cos^2 \alpha - \sin^2 \alpha) = \cos^2 \alpha - \sin^2 \alpha \]

Now, recall that \( \cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha \). Thus, we have:

\[ \cos^4 \alpha - \sin^4 \alpha - \cos 2\alpha = (\cos^2 \alpha - \sin^2 \alpha) - (\cos^2 \alpha - \sin^2 \alpha) = 0 \]

Since both the numerator and denominator approach 0 as \( x \to 4 \), we have a \( \frac{0}{0} \) form.

**Step 2: Apply L'Hôpital's Rule**

According to L'Hôpital's Rule, we differentiate the numerator and the denominator separately:

- The derivative of the numerator \( (\cos \alpha)^x - (\sin \alpha)^x - \cos 2\alpha \):

Using the formula for the derivative of \( a^x \), which is \( a^x \ln a \):

\[ \frac{d}{dx}[(\cos \alpha)^x] = (\cos \alpha)^x \ln(\cos \alpha) \]
\[ \frac{d}{dx}[(\sin \alpha)^x] = (\sin \alpha)^x \ln(\sin \alpha) \]
\[ \frac{d}{dx}[-\cos 2\alpha] = 0 \]

Thus, the derivative of the numerator is:

\[ (\cos \alpha)^x \ln(\cos \alpha) - (\sin \alpha)^x \ln(\sin \alpha) \]

- The derivative of the denominator \( x - 4 \) is simply \( 1 \).

Now we can rewrite the limit:

\[ \lim_{x \to 4} \left[ (\cos \alpha)^x \ln(\cos \alpha) - (\sin \alpha)^x \ln(\sin \alpha) \right] \]

**Step 3: Substitute \( x = 4 \)**

Now we substitute \( x = 4 \):

\[ \lim_{x \to 4} \left[ (\cos \alpha)^4 \ln(\cos \alpha) - (\sin \alpha)^4 \ln(\sin \alpha) \right] \]

This gives us:

\[ (\cos \alpha)^4 \ln(\cos \alpha) - (\sin \alpha)^4 \ln(\sin \alpha) \]

**Final Result:**

The value of the limit is:

\[ (\cos \alpha)^4 \ln(\cos \alpha) - (\sin \alpha)^4 \ln(\sin \alpha) \]

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