We are tasked with evaluating the given limit:

$\lim_{x \to 0} f(x) = \lim_{x \to 0} \left( \frac{\tan x}{x} + \frac{\tan 2x}{\tan 2x} \right).$

Step 1: Simplify the components

1. First term:
   As $x \to 0$, the limit $\frac{\tan x}{x} \to 1$, because $\tan x \sim x$ near zero.

2. Second term:
   Simplify $\frac{\tan 2x}{\tan 2x}$: \frac{\tan 2x}{\tan 2x} = 1 \quad \text{(for all x \neq 0)}.

Step 2: Evaluate the entire limit

Combining these results: $\lim_{x \to 0} f(x) = \lim_{x \to 0} \left( \frac{\tan x}{x} + \frac{\tan 2x}{\tan 2x} \right) = 1 + 1 = 2.$

Final Answer:

$\boxed{2}$

Let me know if you'd like further clarification or details!

Related Questions:

1. What is the derivation of $\lim_{x \to 0} \frac{\tan x}{x} = 1$?
2. How would this limit change if $\tan 2x$ in the second term were replaced with $\sin 2x$?
3. What happens if $x \to \pi$ instead of $x \to 0$ in this limit?
4. How is the small-angle approximation for $\tan x \sim x$ derived?
5. Can L'Hôpital's rule be used to verify the result for the first term?

Tip:

For limits involving trigonometric functions, the small-angle approximations $\sin x \sim x$ and $\tan x \sim x$ are extremely useful!