Question

$\lim_{\theta\rightarrow 0}\frac{\theta \sin 2\theta}{1-\cos 2\theta} =$

• A. 1
• B. $\frac{-1}{2}$
• C. -1
• D. $\frac{1}{2}$
• E. 0

Hint:

Small angle approximations $(\sin \theta \approx \theta, \cos \theta \approx 1 - \frac{\theta^2}{2})$ can simplify limits involving trigonometric terms.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to evaluate a limit involving trigonometric functions as the variable approaches zero. The expression is in the indeterminate form \( \frac{0}{0} \), so we must simplify it using trigonometric identities or L'Hôpital's Rule.
Step 2: Key Formula or Approach:
We will use the following trigonometric identities:
1. Double-angle for sine: \( \sin 2\theta = 2 \sin \theta \cos \theta \)
2. Half-angle identity for cosine (rearranged): \( 1 - \cos 2\theta = 2 \sin^2 \theta \)
And the fundamental trigonometric limit:
3. \( \lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1 \)
Step 3: Detailed Explanation:
Let's substitute the identities into the limit expression.
\[ \lim_{\theta \to 0} \frac{\theta (2 \sin \theta \cos \theta)}{2 \sin^2 \theta} \] Cancel the common factor of 2:
\[ \lim_{\theta \to 0} \frac{\theta \sin \theta \cos \theta}{\sin^2 \theta} \] Assuming \( \theta \neq 0 \), we can cancel one factor of \( \sin \theta \) from the numerator and denominator:
\[ \lim_{\theta \to 0} \frac{\theta \cos \theta}{\sin \theta} \] Rearrange the expression to use the standard limit:
\[ \lim_{\theta \to 0} \left( \frac{\theta}{\sin \theta} \right) \cdot \cos \theta \] We can separate this into the product of two limits:
\[ \left( \lim_{\theta \to 0} \frac{\theta}{\sin \theta} \right) \cdot \left( \lim_{\theta \to 0} \cos \theta \right) \] The first limit is the reciprocal of the standard limit: \( \lim_{\theta \to 0} \frac{1}{\frac{\sin \theta}{\theta}} = \frac{1}{1} = 1 \).
The second limit is: \( \cos(0) = 1 \).
So, the final result is:
\[ 1 \cdot 1 = 1 \] Step 4: Final Answer:
The value of the limit is 1.