To find two functions $f(x)$ and $g(x)$ such that:

1. $\lim_{x \to 0} f(x)$ does not exist,
2. $\lim_{x \to 0} g(x)$ does not exist,
3. $\lim_{x \to 0} \frac{f(x)}{g(x)}$ exists,

we can consider oscillatory functions that have undefined limits at $x = 0$, but their ratio produces a well-defined limit.

One good pair of functions that meets these criteria is:

• $f(x) = \sin\left(\frac{1}{x}\right)$
• $g(x) = \frac{1}{x} \sin\left(\frac{1}{x}\right)$

Why these functions work:

1. Limit of $f(x) = \sin\left(\frac{1}{x}\right)$ as $x \to 0$:

   • This function oscillates between -1 and 1 as $x \to 0$, so the limit does not exist.

2. Limit of $g(x) = \frac{1}{x} \sin\left(\frac{1}{x}\right)$ as $x \to 0$:

   • This also oscillates, and grows unbounded as $x \to 0$, so the limit does not exist.

3. Limit of $\frac{f(x)}{g(x)}$ as $x \to 0$: $\frac{f(x)}{g(x)} = \frac{\sin\left(\frac{1}{x}\right)}{\frac{1}{x} \sin\left(\frac{1}{x}\right)} = x$

   • The limit of $x$ as $x \to 0$ is clearly 0, so the limit of the ratio exists and is 0.

Thus, we have found two functions where the individual limits do not exist, but their ratio has a well-defined limit of 0 as $x \to 0$.

Would you like further explanation on this example? Here are 5 related questions:

1. Can we choose other pairs of oscillatory functions for similar results?
2. What happens if we use different scaling factors for $f(x)$ and $g(x)$?
3. How can we prove that these limits do not exist rigorously?
4. Could we use piecewise-defined functions to achieve similar behavior?
5. How would these limits change if we approached a different point other than zero?

Tip: When solving limit problems, particularly with undefined behaviors, try looking for patterns in oscillatory functions or division by zero scenarios to explore more interesting cases.