We need to evaluate the limit:

$\lim_{x \to 0} \frac{(\cos x)^3}{x^2}$

Step 1: Check Direct Substitution

Substituting $x = 0$:

$\frac{(\cos 0)^3}{0^2} = \frac{1^3}{0} = \frac{1}{0}$

which is undefined, so we must use another method to evaluate the limit.

Step 2: Use Taylor Series Expansion for $\cos x$

The Taylor series expansion of $\cos x$ at $x = 0$ is:

$\cos x = 1 - \frac{x^2}{2} + O(x^4)$

Raising it to the third power:

$(\cos x)^3 = \left(1 - \frac{x^2}{2} + O(x^4) \right)^3$

Using the binomial approximation:

$(\cos x)^3 \approx 1 - \frac{3x^2}{2} + O(x^4)$

Step 3: Compute the Limit

$\frac{(\cos x)^3}{x^2} = \frac{1 - \frac{3x^2}{2} + O(x^4)}{x^2}$

Splitting the terms:

$\frac{1}{x^2} - \frac{3}{2} + O(x^2)$

Since $\frac{1}{x^2} \to \infty$ as $x \to 0$, the entire expression goes to infinity.

Conclusion:

$\lim_{x \to 0} \frac{(\cos x)^3}{x^2} = \infty$

This means the function has a vertical asymptote at $x = 0$.

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Would you like a more detailed explanation? Let me know if you have any questions!

Related Questions:

1. What is the Taylor series expansion of $\cos x$ up to $x^6$?
2. How do you evaluate limits involving trigonometric functions and indeterminate forms?
3. What is the behavior of $\frac{\sin x}{x}$ as $x \to 0$?
4. How do you use L'Hôpital’s Rule to evaluate limits?
5. What are common limits involving trigonometric functions?

Tip: When dealing with limits that involve trigonometric functions, Taylor series expansions can be very useful!