The limit shown in the image is:

$\lim_{{x \to \frac{\pi}{2}}} (\sec x - \tan x)$

To solve this limit as $x \to \frac{\pi}{2}$, let's break it down:

1. Secant and Tangent Behavior at $x = \frac{\pi}{2}$:

   • $\sec x = \frac{1}{\cos x}$
   • $\tan x = \frac{\sin x}{\cos x}$

   As $x \to \frac{\pi}{2}$, $\cos x \to 0$, causing both $\sec x$ and $\tan x$ to approach infinity.

2. Analyzing the expression: $\sec x - \tan x = \frac{1}{\cos x} - \frac{\sin x}{\cos x}$ This simplifies to: $\frac{1 - \sin x}{\cos x}$

   Now, as $x \to \frac{\pi}{2}$:

   • $\sin \frac{\pi}{2} = 1$
   • $\cos \frac{\pi}{2} = 0$

   Therefore, the numerator $1 - \sin x \to 0$ and the denominator $\cos x \to 0$, leading to an indeterminate form $\frac{0}{0}$.

3. L'Hopital's Rule: To resolve this indeterminate form, apply L'Hopital's Rule by differentiating the numerator and denominator.

   • The derivative of $1 - \sin x$ is $-\cos x$.
   • The derivative of $\cos x$ is $-\sin x$.

   So, applying L'Hopital's Rule gives us: $\lim_{{x \to \frac{\pi}{2}}} \frac{-\cos x}{-\sin x} = \lim_{{x \to \frac{\pi}{2}}} \frac{\cos x}{\sin x}$ As $x \to \frac{\pi}{2}$, $\cos \frac{\pi}{2} = 0$ and $\sin \frac{\pi}{2} = 1$, so the limit evaluates to: $0$

Thus, the value of the limit is:

$\lim_{{x \to \frac{\pi}{2}}} (\sec x - \tan x) = 0$

Would you like a more detailed explanation or have any questions?

Here are some related questions:

1. How is L'Hopital's Rule applied in indeterminate forms like $\frac{0}{0}$?
2. What is the behavior of trigonometric functions near $\frac{\pi}{2}$?
3. Can the limit be solved using series expansion instead of L'Hopital's Rule?
4. Why does the difference of secant and tangent behave differently from their individual limits?
5. How do other limits involving secant and tangent behave near their undefined points?

Tip: When evaluating limits of trigonometric functions near critical points like $\frac{\pi}{2}$, understanding the behavior of sine and cosine is crucial, as they determine how secant and tangent behave.