Let `lim_(x to 0) ("sin" 2X)/(tan ((x)/(2))) = L,` and `lim_(x to 0) (e^(2x) - 1)/(x) = L_(2)` then the value of `L_(1)L_(2)` is

Options

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

Answer

To solve the given problem, we need to evaluate two limits and then find the product of those limits.

### Step 1: Evaluate \( L_1 = \lim_{x \to 0} \frac{\sin(2x)}{\tan\left(\frac{x}{2}\right)} \)

We can use the properties of limits to simplify this expression. We know that:

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1 \quad \text{and} \quad \lim_{x \to 0} \frac{\tan x}{x} = 1 \]

To manipulate our limit, we can rewrite it:

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

Now, we can separate the limit into two parts:

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

### Step 2: Evaluate \( \lim_{x \to 0} \frac{\sin(2x)}{2x} \)

Using the limit property:

\[ \lim_{x \to 0} \frac{\sin(2x)}{2x} = 1 \]

### Step 3: Evaluate \( \lim_{x \to 0} \frac{2x}{\tan\left(\frac{x}{2}\right)} \)

Using the limit property:

\[ \lim_{x \to 0} \frac{\tan\left(\frac{x}{2}\right)}{\frac{x}{2}} = 1 \implies \lim_{x \to 0} \frac{2x}{\tan\left(\frac{x}{2}\right)} = 1 \]

### Step 4: Combine the results for \( L_1 \)

Now we can combine the results:

\[ L_1 = 1 \cdot 1 = 1 \]

### Step 5: Evaluate \( L_2 = \lim_{x \to 0} \frac{e^{2x} - 1}{x} \)

We can use the limit property:

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

To evaluate \( L_2 \):

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

### Step 6: Calculate \( L_1 \cdot L_2 \)

Now we can find the product of \( L_1 \) and \( L_2 \):

\[ L_1 \cdot L_2 = 1 \cdot 2 = 2 \]

### Final Answer

Thus, the value of \( L_1 L_2 \) is \( 2 \).

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