Let `lim_(xrarr0)(sin2x)/(tan((x)/(k)))=L_(1) and lim_(xrarr0)(e^(2x)-1)/(x)=L_(2),` and the value of `L_(1) L_(2)` is 8, then k is

Options

A. 4
B. 8
C. 6
D. 2

Answer

To solve the problem, we need to calculate the limits \( L_1 \) and \( L_2 \), and then find the value of \( k \) such that \( L_1 \cdot L_2 = 8 \).

### Step 1: Calculate \( L_1 \)

We start with the limit:

\[ L_1 = \lim_{x \to 0} \frac{\sin(2x)}{\tan\left(\frac{x}{k}\right)} \]

Using the identity \( \tan(x) \approx x \) as \( x \to 0 \), we can rewrite \( \tan\left(\frac{x}{k}\right) \):

\[ \tan\left(\frac{x}{k}\right) \approx \frac{x}{k} \]

Thus, we can rewrite \( L_1 \):

\[ L_1 = \lim_{x \to 0} \frac{\sin(2x)}{\frac{x}{k}} = k \cdot \lim_{x \to 0} \frac{\sin(2x)}{x} \]

Now, we know that \( \lim_{x \to 0} \frac{\sin(2x)}{2x} = 1 \), so:

\[ \lim_{x \to 0} \frac{\sin(2x)}{x} = 2 \cdot \lim_{x \to 0} \frac{\sin(2x)}{2x} = 2 \]

Therefore:

\[ L_1 = k \cdot 2 \]

### Step 2: Calculate \( L_2 \)

Next, we calculate \( L_2 \):

\[ L_2 = \lim_{x \to 0} \frac{e^{2x} - 1}{x} \]

As \( x \to 0 \), this limit is of the form \( \frac{0}{0} \), so we can apply L'Hôpital's Rule:

\[ L_2 = \lim_{x \to 0} \frac{d}{dx}(e^{2x} - 1) \bigg/ \frac{d}{dx}(x) = \lim_{x \to 0} \frac{2e^{2x}}{1} \]

Evaluating this limit as \( x \to 0 \):

\[ L_2 = 2e^{0} = 2 \]

### Step 3: Find \( k \)

Now we have:

\[ L_1 = 2k \quad \text{and} \quad L_2 = 2 \]

We know from the problem statement that:

\[ L_1 \cdot L_2 = 8 \]

Substituting the values of \( L_1 \) and \( L_2 \):

\[ (2k) \cdot 2 = 8 \]

This simplifies to:

\[ 4k = 8 \]

Dividing both sides by 4 gives:

\[ k = 2 \]

### Final Answer

Thus, the value of \( k \) is:

\[ \boxed{2} \]