$lim_{x\rightarrow0}\frac{sin~4x}{sin~2x}=$

Q: $lim_{x\rightarrow0}\frac{sin~4x}{sin~2x}=$

To solve the limit problem $ \lim_{x \rightarrow 0} \frac{\sin 4x}{\sin 2x} $, we can use the standard limit property: \[ \lim_{x \rightarrow 0} \frac{\sin kx}{kx} = 1 \] for any constant $ k $. This property helps us simplify trigonometric limits. Let's apply this property to the given expression: 1. Rewrite the expression using the limit property: \[ \lim_{x \rightarrow 0} \frac{\sin 4x}{\sin 2x} = \lim_{x \rightarrow 0} \frac{\sin 4x}{4x} \cdot \frac{2x}{\sin 2x} \cdot \frac{4}{2} \] 2. Apply the limit property: \[ \lim_{x \rightarrow 0} \frac{\sin 4x}{4x} = 1 \quad \text{and} \quad \lim_{x \rightarrow 0} \frac{2x}{\sin 2x} = 1 \] 3. Substitute these results back into the expression: \[ \lim_{x \rightarrow 0} \frac{\sin 4x}{\sin 2x} = 1 \cdot 1 \cdot 2 = 2 \] Thus, the correct answer is Option 4: 2. **Explanation for why other options are incorrect:** - **Option 1 (1):** This would imply that the limit simplifies to 1, which is incorrect because the factor of 2 from the expression $\frac{4}{2}$ is not accounted for. - **Option 2 (-1):** This option is incorrect because the limit does not involve any negative factors that would result in a negative value. - **Option 3 (-2):** Similar to Option 2, there are no negative factors in the expression that would result in a negative value, making this option incorrect. Therefore, the correct answer is indeed Option 4: 2, as derived from the application of the standard trigonometric limit properties.

Question

Accepted Answer

To solve the limit problem $ \lim_{x \rightarrow 0} \frac{\sin 4x}{\sin 2x} $, we can use the standard limit property:

$$
\lim_{x \rightarrow 0} \frac{\sin kx}{kx} = 1
$$

for any constant $ k $. This property helps us simplify trigonometric limits.

Let's apply this property to the given expression:

1. Rewrite the expression using the limit property:
   $$
   \lim_{x \rightarrow 0} \frac{\sin 4x}{\sin 2x} = \lim_{x \rightarrow 0} \frac{\sin 4x}{4x} \cdot \frac{2x}{\sin 2x} \cdot \frac{4}{2}
   $$

2. Apply the limit property:
   $$
   \lim_{x \rightarrow 0} \frac{\sin 4x}{4x} = 1 \quad \text{and} \quad \lim_{x \rightarrow 0} \frac{2x}{\sin 2x} = 1
   $$

3. Substitute these results back into the expression:
   $$
   \lim_{x \rightarrow 0} \frac{\sin 4x}{\sin 2x} = 1 \cdot 1 \cdot 2 = 2
   $$

Thus, the correct answer is Option 4: 2.

**Explanation for why other options are incorrect:**

- **Option 1 (1):** This would imply that the limit simplifies to 1, which is incorrect because the factor of 2 from the expression $\frac{4}{2}$ is not accounted for.

- **Option 2 (-1):** This option is incorrect because the limit does not involve any negative factors that would result in a negative value.

- **Option 3 (-2):** Similar to Option 2, there are no negative factors in the expression that would result in a negative value, making this option incorrect.

Therefore, the correct answer is indeed Option 4: 2, as derived from the application of the standard trigonometric limit properties.