Let's go through each limit problem step by step.

(a) $\lim_{{x \to 0}} \frac{{\tan x}}{x}$

To solve this, we use the fact that as $x \to 0$, $\tan x \approx x$. Therefore: $\lim_{{x \to 0}} \frac{{\tan x}}{x} = \lim_{{x \to 0}} \frac{{x}}{x} = 1$

Thus, the answer is: $\boxed{1}$

(b) $\lim_{{\theta \to 0}} \frac{{\sin 2\theta}}{\theta}$

Rewrite $\sin 2\theta$ using a substitution. Let $u = 2\theta$. Then, as $\theta \to 0$, $u \to 0$ as well. The expression becomes: $\lim_{{\theta \to 0}} \frac{{\sin 2\theta}}{\theta} = \lim_{{u \to 0}} \frac{{\sin u}}{\frac{u}{2}}$ $= 2 \cdot \lim_{{u \to 0}} \frac{{\sin u}}{u}$

We know from the standard limit that $\lim_{{u \to 0}} \frac{{\sin u}}{u} = 1$. So: $= 2 \cdot 1 = 2$

Thus, the answer is: $\boxed{2}$

(c) $\lim_{{x \to 0}} \frac{{\sin 3x}}{\sin 5x}$

To solve this, we use the fact that $\lim_{{x \to 0}} \frac{{\sin ax}}{ax} = 1$. Rewrite the limit as: $\lim_{{x \to 0}} \frac{{\sin 3x}}{\sin 5x} = \lim_{{x \to 0}} \frac{{\sin 3x}}{3x} \cdot \frac{{5x}}{\sin 5x} \cdot \frac{3}{5}$

Applying the standard limits: $\lim_{{x \to 0}} \frac{{\sin 3x}}{3x} = 1 \quad \text{and} \quad \lim_{{x \to 0}} \frac{{5x}}{\sin 5x} = 1$

Therefore: $= 1 \cdot 1 \cdot \frac{3}{5} = \frac{3}{5}$

Thus, the answer is: $\boxed{\frac{3}{5}}$

Summary

• (a) $\boxed{1}$
• (b) $\boxed{2}$
• (c) $\boxed{\frac{3}{5}}$

Would you like more details on any of these steps or have any further questions?

Here are five related questions to expand your understanding:

1. What is the limit of $\frac{\sin x}{x}$ as $x \to 0$?
2. How do you find the limit of $\frac{\tan 2x}{x}$ as $x \to 0$?
3. Can you explain why $\lim_{{x \to 0}} \frac{{\sin ax}}{x} = a$?
4. How can L'Hôpital's Rule be used to solve trigonometric limits?
5. What is the limit of $\frac{\cos x - 1}{x}$ as $x \to 0$?

Tip: Always remember that standard trigonometric limits, such as $\lim_{{x \to 0}} \frac{{\sin x}}{x} = 1$, are fundamental and frequently used in calculus to evaluate other limits.