To solve this problem, we need to find an example where the limit of the product of two functions $f(x)$ and $g(x)$ as $x \to a$ exists, even though the individual limits $\lim_{x \to a} f(x)$ and $\lim_{x \to a} g(x)$ do not exist.

Example:

Consider the functions: $f(x) = \sin\left(\frac{1}{x}\right) \quad \text{and} \quad g(x) = \frac{1}{x}$ as $x \to 0$.

1. Examining $f(x) = \sin\left(\frac{1}{x}\right)$:

   • As $x \to 0$, the argument $\frac{1}{x}$ oscillates between positive and negative infinity, causing $\sin\left(\frac{1}{x}\right)$ to oscillate between -1 and 1.
   • Therefore, $\lim_{x \to 0} \sin\left(\frac{1}{x}\right)$ does not exist because of the oscillations.

2. Examining $g(x) = \frac{1}{x}$:

   • As $x \to 0$, $\frac{1}{x}$ diverges to either $+\infty$ or $-\infty$, depending on the direction from which $x$ approaches 0.
   • Thus, $\lim_{x \to 0} \frac{1}{x}$ also does not exist.

3. Examining the product $f(x) \cdot g(x) = \sin\left(\frac{1}{x}\right) \cdot \frac{1}{x}$:

   • We analyze $\lim_{x \to 0} \sin\left(\frac{1}{x}\right) \cdot \frac{1}{x}$.
   • Using the fact that $-1 \leq \sin\left(\frac{1}{x}\right) \leq 1$, we can bound the product: $-\frac{1}{x} \leq \sin\left(\frac{1}{x}\right) \cdot \frac{1}{x} \leq \frac{1}{x}$
   • However, as $x \to 0$, this expression behaves more gently. Specifically: [ \lim_{x \to