To examine the continuity of the given function $f(x)$ at $x = 0$, we need to check whether the following three conditions for continuity are satisfied:

1. Existence of $f(0)$:
   The function is defined at $x = 0$, and from the definition,
   $f(0) = 0.$

2. Existence of the limit $\lim_{x \to 0} f(x)$:
   We need to evaluate the limit of $f(x)$ as $x \to 0$. For $x \neq 0$, the function is given by
   $f(x) = \sin\left(\frac{1}{x}\right).$ We must now check if this limit exists. As $x \to 0$, the value $\frac{1}{x}$ becomes unbounded, which means the argument of the sine function oscillates between $-\infty$ and $+\infty$. Since $\sin\left(\frac{1}{x}\right)$ oscillates between $-1$ and $1$ as $x \to 0$, the limit does not settle on any single value. Hence,
   $\lim_{x \to 0} \sin\left(\frac{1}{x}\right) \quad \text{does not exist}.$

3. Equality of the limit and $f(0)$:
   Even though $f(0) = 0$, the limit $\lim_{x \to 0} f(x)$ does not exist, so this condition cannot be satisfied.

Conclusion:

The function $f(x)$ is not continuous at $x = 0$ because the limit $\lim_{x \to 0} f(x)$ does not exist.

Would you like more details or have any questions about this?

Here are five related questions for further exploration:

1. What is the formal definition of continuity at a point?
2. How does the oscillatory behavior of $\sin\left(\frac{1}{x}\right)$ affect limits?
3. Can a function be discontinuous at a point but continuous elsewhere?
4. What types of discontinuities exist, and how do they differ?
5. How can we construct functions with similar behaviors around points of discontinuity?

Tip: When dealing with oscillatory functions, it's helpful to analyze their behavior by looking at upper and lower bounds, as this can provide insight into the behavior of limits.