91Ó°ÊÓ

Indeterminate forms occur when evaluating a limit leads to expressions like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \). These forms are not straightforwardly solvable as they do not lead to a clear outcome without further manipulation.
In the given problem, as \( x \to \frac{\pi}{2}^- \), the term \( x - \frac{\pi}{2} \) tends to \( 0^- \), and \( \sec x = \frac{1}{\cos x} \) leads to \(-\infty\) because \( \cos x \to 0^+ \). The product of these approaching values creates an ambiguous expression \( 0 \times -\infty \).
To deal with this, we rewrite it into a form that's suitable for l'Hôpital's rule. By converting the product \( 0 \times -\infty \) into a ratio \( \frac{0}{0} \), we prepare the expression for l'Hôpital's intervention.

The concept of a limit is foundational in calculus, describing the behavior of a function as it approaches a particular point. In this context, limits help us understand the value that \( x \) approaches.
When applying limits to expressions like the one given, we notice how both the numerator and denominator approach zero, forming a classic \( \frac{0}{0} \) scenario. Limits like these require special techniques, such as l'Hôpital's rule, to resolve.
This rule states that if the limit \( \lim_{x \to c} \frac{f(x)}{g(x)} \) results in an indeterminate form, then \( \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \), assuming this latter limit exists. Thus, through this process, the indeterminate form becomes manageable, guiding us toward the function's behavior at \( x = \frac{\pi}{2} \).

Derivatives play a crucial part in making complex limits solvable, especially when dealing with indeterminate forms. They measure how a function changes as its input changes.
In our exercise, when using l'Hôpital's rule, derivatives come into play. We first differentiate the numerator \( x - \frac{\pi}{2} \), which gives us \( 1 \). Next, we differentiate the denominator \( \cos x \), yielding \( -\sin x \). These derivatives are critical in transforming our \( \frac{0}{0} \) form into a simpler limit.
We then evaluate \( \lim _{x \to \frac{\pi}{2}^-} \frac{1}{-\sin x} \). As \( x \to \frac{\pi}{2}^- \), \( -\sin x \) approaches \( -1 \), making our limit \( \frac{1}{-1} = -1 \). The derivative process allows us to convert an initially unsolvable problem into a direct evaluation, demystifying complex limit calculations.