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 find the value of \( k \) given the limits \( L_1 \) and \( L_2 \) and the condition that \( L_1 L_2 = 8 \).

### Step-by-Step Solution:

1. **Calculate \( L_1 \)**:
\[ L_1 = \lim_{x \to 0} \frac{\sin(2x)}{\tan\left(\frac{x}{k}\right)} \]
We know that \( \sin(2x) \) can be rewritten using the limit property \( \lim_{x \to 0} \frac{\sin x}{x} = 1 \). Thus, we can manipulate the limit:
\[ L_1 = \lim_{x \to 0} \frac{\sin(2x)}{2x} \cdot \frac{2x}{\tan\left(\frac{x}{k}\right)} \cdot \frac{1}{k} \]
This gives us:
\[ L_1 = \lim_{x \to 0} \left( \frac{\sin(2x)}{2x} \cdot \frac{2x}{\tan\left(\frac{x}{k}\right)} \cdot k \right) \]
As \( x \to 0 \), \( \frac{\sin(2x)}{2x} \to 1 \) and \( \frac{2x}{\tan\left(\frac{x}{k}\right)} \to 1 \) (since \( \tan\left(\frac{x}{k}\right) \sim \frac{x}{k} \) as \( x \to 0 \)). Therefore:
\[ L_1 = 2k \]

2. **Calculate \( L_2 \)**:
\[ L_2 = \lim_{x \to 0} \frac{e^{2x} - 1}{x} \]
We can use the standard limit \( \lim_{x \to 0} \frac{e^{x} - 1}{x} = 1 \). Thus:
\[ L_2 = \lim_{x \to 0} \frac{e^{2x} - 1}{2x} \cdot 2 \]
This gives:
\[ L_2 = 2 \cdot 1 = 2 \]

3. **Combine \( L_1 \) and \( L_2 \)**:
Given that \( L_1 L_2 = 8 \):
\[ (2k)(2) = 8 \]
Simplifying this:
\[ 4k = 8 \]
Therefore:
\[ k = \frac{8}{4} = 2 \]

### Final Answer:
The value of \( k \) is \( 2 \).

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