To solve the limit problem in the image:

$\lim_{\theta \to 0} \frac{\sin \theta}{\theta + \tan \theta}$

we'll use known limit properties and trigonometric approximations around $\theta = 0$.

Step-by-Step Solution

1. Rewrite $\tan \theta$ in terms of $\sin \theta$ and $\cos \theta$: $\tan \theta = \frac{\sin \theta}{\cos \theta}$ Thus, the expression becomes: $\lim_{\theta \to 0} \frac{\sin \theta}{\theta + \frac{\sin \theta}{\cos \theta}}$

2. Combine terms in the denominator: Multiply the numerator and the denominator by $\cos \theta$ to simplify: $\lim_{\theta \to 0} \frac{\sin \theta \cdot \cos \theta}{\theta \cos \theta + \sin \theta}$

3. Apply limits separately to each part of the fraction: Using the small-angle approximations around $\theta = 0$, we know:

   • $\sin \theta \approx \theta$
   • $\cos \theta \approx 1$

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

4. Simplify the expression: $\lim_{\theta \to 0} \frac{\theta}{2\theta} = \lim_{\theta \to 0} \frac{1}{2} = \frac{1}{2}$

Conclusion

The limit is: $\lim_{\theta \to 0} \frac{\sin \theta}{\theta + \tan \theta} = \frac{1}{2}$

Would you like a deeper breakdown of any step or have additional questions?

Here are five related questions to extend your understanding:

1. How would the limit change if the denominator were $\theta^2 + \tan \theta$?
2. Can L'Hôpital's Rule be applied to solve this limit?
3. Why is $\sin \theta \approx \theta$ a valid approximation near $\theta = 0$?
4. How would you approach this limit if $\theta \to \pi$ instead of $\theta \to 0$?
5. What happens to the limit if $\theta$ approaches zero from the negative side?

Tip: Small-angle approximations are very useful for trigonometric limits near zero.