To solve the given limit problem, we need to carefully analyze the expression:

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

Let’s break it down step-by-step.

Step 1: Simplify the Expression Inside the Parentheses

Rewrite $\frac{1}{\sin \theta} - \frac{1}{\tan \theta}$: $\frac{1}{\sin \theta} - \frac{1}{\tan \theta} = \frac{1}{\sin \theta} - \frac{\cos \theta}{\sin \theta}$ This simplifies to: $= \frac{1 - \cos \theta}{\sin \theta}$

Step 2: Substitute This Back into the Limit

The expression now becomes: $\lim_{\theta \to 0} \frac{1}{\theta} \cdot \frac{1 - \cos \theta}{\sin \theta} = \lim_{\theta \to 0} \frac{1 - \cos \theta}{\theta \sin \theta}$

Step 3: Apply L'Hôpital's Rule if Necessary

Since $\frac{1 - \cos \theta}{\theta \sin \theta}$ is an indeterminate form $\frac{0}{0}$ as $\theta \to 0$, we can use L'Hôpital's Rule by differentiating the numerator and the denominator with respect to $\theta$.

1. Differentiate the numerator $1 - \cos \theta$: $\frac{d}{d\theta} (1 - \cos \theta) = \sin \theta$

2. Differentiate the denominator $\theta \sin \theta$: $\frac{d}{d\theta} (\theta \sin \theta) = \sin \theta + \theta \cos \theta$

Thus, the limit becomes: $\lim_{\theta \to 0} \frac{\sin \theta}{\sin \theta + \theta \cos \theta}$

Step 4: Substitute $\theta = 0$ in the Simplified Form

As $\theta \to 0$, $\sin \theta \to 0$ and $\theta \cos \theta \to 0$, so the expression simplifies to: $\lim_{\theta \to 0} \frac{\sin \theta}{\sin \theta + \theta \cos \theta} = \frac{0}{0 + 0} = 0$

Therefore, the final answer is:

$\boxed{0}$

Would you like further details or explanations on any part of this process?

Related Questions:

1. How does L'Hôpital's Rule apply to trigonometric functions in limits?
2. Why do we get an indeterminate form in this limit problem?
3. What are common techniques for simplifying limits involving trigonometric functions?
4. How can we use Taylor series to approximate trigonometric functions near zero?
5. What other methods exist for solving limits if L'Hôpital’s Rule is not applicable?

Tip:

When working with trigonometric limits as $\theta \to 0$, remember that using small-angle approximations (like $\sin \theta \approx \theta$ and $1 - \cos \theta \approx \frac{\theta^2}{2}$) can often simplify calculations significantly.