To evaluate the limit $\lim_{x \to 0} \frac{(\cos x)^3}{x^2}$ using L'Hopital's Rule, let's first check if we have a $\frac{0}{0}$ indeterminate form when substituting $x = 0$.

Substituting $x = 0$:

• $\cos(0) = 1$, so $(\cos(0))^3 = 1^3 = 1$.
• $x^2 = 0^2 = 0$.

Thus, we get the indeterminate form $\frac{1}{0}$, which is not in the form of $\frac{0}{0}$, but rather $\frac{1}{0}$. This means the limit does not exist in the conventional sense (it tends to infinity).

Since the limit tends to infinity, we conclude that:

$\lim_{x \to 0} \frac{(\cos x)^3}{x^2} \text{ does not exist and approaches infinity.}$

Would you like to explore any other techniques or concepts related to this? Here are a few related questions you could consider:

1. What is the behavior of the limit when the numerator is a higher power of $\cos x$?
2. How can you apply L'Hopital's Rule in cases of indeterminate forms like $\frac{0}{0}$?
3. How would the limit change if we considered a different function in place of $\cos x$?
4. What happens if we take the limit from one side (left or right) of zero?
5. How do we deal with limits of functions involving trigonometric identities as $x \to 0$?

Tip: Always check the type of indeterminate form before applying L'Hopital's Rule; it only works for $0/0$ or $\infty/\infty$.