If `lim_(xto0)(log_(e)cot((pi)/4-K_(1)x))/(tanK_(2)x)=1`, then

Options

A. `K_(1)=K_(2)`
B. `2K_(1)=K_(2)`
C. `K_(1)=2K_(2)`
D. `K_(1)=4K_(2)`

Answer

To solve the limit problem given by

\[ \lim_{x \to 0} \frac{\log_e \cot\left(\frac{\pi}{4} - K_1 x\right)}{\tan(K_2 x)} = 1, \]

we will follow these steps:

### Step 1: Identify the form of the limit

As \( x \to 0 \):
- \(\cot\left(\frac{\pi}{4} - K_1 x\right) \to \cot\left(\frac{\pi}{4}\right) = 1\), thus \(\log_e \cot\left(\frac{\pi}{4} - K_1 x\right) \to \log_e(1) = 0\).
- \(\tan(K_2 x) \to \tan(0) = 0\).

This gives us a \( \frac{0}{0} \) indeterminate form, which allows us to apply L'Hôpital's Rule.

### Step 2: Apply L'Hôpital's Rule

By L'Hôpital's Rule, we differentiate the numerator and denominator:

\[ \text{Numerator: } \frac{d}{dx} \left( \log_e \cot\left(\frac{\pi}{4} - K_1 x\right) \right) = \frac{-K_1}{\cot\left(\frac{\pi}{4} - K_1 x\right) \cdot \sin^2\left(\frac{\pi}{4} - K_1 x\right)}. \]

Using the chain rule, the derivative of \(\cot\) gives us:

\[ \frac{d}{dx} \left( \cot\left(\frac{\pi}{4} - K_1 x\right) \right) = -K_1 \cdot \csc^2\left(\frac{\pi}{4} - K_1 x\right). \]

Thus, the derivative of the numerator becomes:

\[ \frac{-K_1 \cdot \csc^2\left(\frac{\pi}{4} - K_1 x\right)}{\cot\left(\frac{\pi}{4} - K_1 x\right)}. \]

For the denominator:

\[ \text{Denominator: } \frac{d}{dx} \left( \tan(K_2 x) \right) = K_2 \sec^2(K_2 x). \]

### Step 3: Rewrite the limit

Now we can rewrite the limit as:

\[ \lim_{x \to 0} \frac{-K_1 \cdot \csc^2\left(\frac{\pi}{4} - K_1 x\right)}{\cot\left(\frac{\pi}{4} - K_1 x\right) \cdot K_2 \sec^2(K_2 x)}. \]

### Step 4: Evaluate the limit

Now substituting \( x = 0 \):

- \(\cot\left(\frac{\pi}{4}\right) = 1\)
- \(\csc^2\left(\frac{\pi}{4}\right) = 2\)
- \(\sec^2(0) = 1\)

Thus, we have:

\[ \lim_{x \to 0} \frac{-K_1 \cdot 2}{1 \cdot K_2} = \frac{-2K_1}{K_2}. \]

Setting this equal to 1 (as given in the problem):

\[ \frac{-2K_1}{K_2} = 1. \]

### Step 5: Solve for the relationship between \(K_1\) and \(K_2\)

From the equation:

\[ -2K_1 = K_2 \implies K_2 = -2K_1. \]

### Final Result

Thus, the relationship between \(K_1\) and \(K_2\) is:

\[ K_2 = -2K_1. \]

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