What is `lim_(xto0)(tanx)/(sin2x)` equal to ?

Options

A. `1//2`
B. 1
C. 2
D. Limit does not exist

Answer

To solve the limit \( \lim_{x \to 0} \frac{\tan x}{\sin 2x} \), we can follow these steps:

### Step 1: Rewrite the limit using known limits
We know that:
\[ \lim_{x \to 0} \frac{\tan x}{x} = 1 \quad \text{and} \quad \lim_{x \to 0} \frac{\sin x}{x} = 1 \]
We can use these limits to rewrite our expression.

### Step 2: Express the limit in terms of \(x\)
We can rewrite the limit as:
\[ \lim_{x \to 0} \frac{\tan x}{\sin 2x} = \lim_{x \to 0} \frac{\tan x}{x} \cdot \frac{x}{\sin 2x} \]
Now, we can also express \( \sin 2x \) in terms of \( x \):
\[ \sin 2x = 2 \sin x \cos x \]
Thus, we can rewrite the limit as:
\[ \lim_{x \to 0} \frac{\tan x}{\sin 2x} = \lim_{x \to 0} \frac{\tan x}{x} \cdot \frac{x}{2 \sin x \cos x} \]

### Step 3: Substitute the known limits
Now we can substitute the known limits:
\[ \lim_{x \to 0} \frac{\tan x}{x} = 1 \quad \text{and} \quad \lim_{x \to 0} \frac{x}{\sin x} = 1 \]
Thus:
\[ \lim_{x \to 0} \frac{x}{\sin 2x} = \lim_{x \to 0} \frac{x}{2 \sin x \cos x} = \frac{1}{2} \cdot \lim_{x \to 0} \frac{x}{\sin x} \cdot \lim_{x \to 0} \frac{1}{\cos x} \]
Since \( \lim_{x \to 0} \frac{x}{\sin x} = 1 \) and \( \lim_{x \to 0} \cos x = 1 \):
\[ \lim_{x \to 0} \frac{x}{\sin 2x} = \frac{1}{2} \cdot 1 \cdot 1 = \frac{1}{2} \]

### Step 4: Combine the results
Thus, we have:
\[ \lim_{x \to 0} \frac{\tan x}{\sin 2x} = 1 \cdot \frac{1}{2} = \frac{1}{2} \]

### Final Answer
Therefore, the limit is:
\[ \lim_{x \to 0} \frac{\tan x}{\sin 2x} = \frac{1}{2} \]

---