`lim_(x to 0) (sin 2x)/x` is equal to:

Options

A. 0
B. 1
C. `1//2`
D. 2

Answer

To solve the limit \( \lim_{x \to 0} \frac{\sin 2x}{x} \), we can use the standard limit property of sine and some algebraic manipulation. Here’s the step-by-step solution:

### Step 1: Rewrite the limit
We start with the limit:
\[ \lim_{x \to 0} \frac{\sin 2x}{x} \]

### Step 2: Introduce a substitution
To simplify the expression, we can introduce a substitution. Let \( h = 2x \). Then as \( x \to 0 \), \( h \to 0 \) as well. We can express \( x \) in terms of \( h \):
\[ x = \frac{h}{2} \]
Now, we rewrite the limit in terms of \( h \):
\[ \lim_{x \to 0} \frac{\sin 2x}{x} = \lim_{h \to 0} \frac{\sin h}{\frac{h}{2}} = \lim_{h \to 0} \frac{2 \sin h}{h} \]

### Step 3: Apply the standard limit
We know from calculus that:
\[ \lim_{h \to 0} \frac{\sin h}{h} = 1 \]
Using this property, we can evaluate our limit:
\[ \lim_{h \to 0} \frac{2 \sin h}{h} = 2 \cdot \lim_{h \to 0} \frac{\sin h}{h} = 2 \cdot 1 = 2 \]

### Conclusion
Thus, we find that:
\[ \lim_{x \to 0} \frac{\sin 2x}{x} = 2 \]

### Final Answer
The limit is equal to \( 2 \).

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