To evaluate the given limit:

$\lim_{\theta \to 0} \frac{1}{\theta} \left(\frac{1}{\sin \theta} - \frac{1}{\tan \theta}\right)$

First, simplify the expression inside the parentheses:

Step 1: Simplify $\frac{1}{\sin \theta} - \frac{1}{\tan \theta}$

$\frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta}$

Therefore: $\frac{1}{\sin \theta} - \frac{1}{\tan \theta} = \frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta} = \frac{1 - \cos \theta}{\sin \theta}$

Step 2: Substitute this back into the limit

$\lim_{\theta \to 0} \frac{1}{\theta} \cdot \frac{1 - \cos \theta}{\sin \theta}$

This can be rewritten as: $\lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$

Step 3: Use approximations for small $\theta$

As $\theta \to 0$:

• $\sin \theta \approx \theta$
• $1 - \cos \theta \approx \frac{\theta^2}{2}$

Substitute these approximations: $\lim_{\theta \to 0} \frac{\frac{\theta^2}{2}}{\theta \cdot \theta} = \lim_{\theta \to 0} \frac{\theta^2 / 2}{\theta^2} = \frac{1}{2}$

Final Answer:

$\lim_{\theta \to 0} \frac{1}{\theta} \left(\frac{1}{\sin \theta} - \frac{1}{\tan \theta}\right) = \frac{1}{2}$

Would you like further clarification or details on any step?

Here are some related questions to consider:

1. How do we derive Taylor series for trigonometric functions?
2. What are other methods to evaluate limits involving trigonometric functions?
3. Why do small-angle approximations work effectively for limits?
4. How do we apply L'Hôpital's Rule in evaluating trigonometric limits?
5. What are some common pitfalls when simplifying trigonometric expressions?

Tip: When handling limits with trigonometric functions as $\theta \to 0$, using series expansions or approximations can simplify the process and yield results quickly.