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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

Show by means of an example that \( \lim_{x \to a} \left[ f(x) g(x) \right] \) may exist even though neither \( \lim_{x \to a} f(x) \) nor \( \lim_{x \to a} g(x) \) exists.

Explore an example showing that the limit of a product can exist even if neither function's individual limit exists. Learn through the functions \( f(x) = \sin(\frac{1}{x}) \) and \( g(x) = \frac{1}{x} \) as \( x \to 0 \), using limit and oscillatory function concepts. Suitable for college-level calculus.

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