If the function $f(x) = \frac{\tan(\tan x) - \sin(\sin x)}{\tan x - \sin x}$ is continuous at $x = 0$, then $f(0)$ is equal to:

Step 1: Apply continuity condition

For a function to be continuous at a point, the function's value at that point must be equal to its limit as $x$ approaches that point. Since $f(x)$ is continuous at $x=0$, we need to find the limit of $f(x)$ as $x \to 0$ to determine $f(0)$.

Step 2: Substitute $y = \tan x$ and $z = \sin x$

To simplify the expression, let's introduce new variables. Let $y = \tan x$ and $z = \sin x$. As $x \to 0$, both $y \to 0$ and $z \to 0$. This substitution helps in recognizing a standard limit form.

Step 3: Rewrite the expression using Taylor series expansion

We use the Taylor series expansions for $\tan y$ and $\sin z$ around $0$. This allows us to approximate the functions with polynomials, which simplifies the limit calculation. We only need terms up to $y^3$ and $z^3$ for this problem.

Step 4: Substitute expansions and simplify

Substitute the Taylor series expansions back into the expression. We then separate the term $(y-z)/(y-z)$ which simplifies to $1$, and handle the remaining cubic terms.

Step 5: Evaluate the limit using $y = x + \frac{x^3}{3} + O(x^5)$ and $z = x - \frac{x^3}{6} + O(x^5)$

Now, substitute back $y = \tan x$ and $z = \sin x$. We use their Taylor expansions around $x=0$: $\tan x = x + \frac{x^3}{3} + O(x^5)$ and $\sin x = x - \frac{x^3}{6} + O(x^5)$. Notice that $y-z = (x + \frac{x^3}{3}) - (x - \frac{x^3}{6}) = \frac{x^3}{3} + \frac{x^3}{6} = \frac{3x^3}{6} = \frac{x^3}{2}$. Also, for small $x$, $y^3 \approx x^3$ and $z^3 \approx x^3$. Substituting these into the expression, we find the limit to be $2$.

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