The problem presented asks for the limit of $\sec(x)$ as $x \to 0$, and includes a graph to support understanding. Here's the breakdown:

Problem:

We need to evaluate: $\lim_{x \to 0} \sec(x)$

Step-by-step solution:

1. The function $\sec(x)$ is defined as: $\sec(x) = \frac{1}{\cos(x)}$
2. As $x \to 0$, we know that $\cos(0) = 1$. Therefore: $\sec(0) = \frac{1}{1} = 1$
3. Since $\sec(x)$ is continuous near $x = 0$, the limit as $x$ approaches 0 from both sides exists and is equal to 1.

Thus, we conclude that: $\lim_{x \to 0} \sec(x) = 1$

Graph interpretation:

The graph in the image shows a curve resembling $y = \sec(x)$, with vertical asymptotes at $x = \pm \frac{\pi}{2}$, indicating that the function becomes undefined at those points. At $x = 0$, the value of the function is 1, as expected.

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Related questions:

1. What happens to $\sec(x)$ as $x \to \frac{\pi}{2}$?
2. How does $\sec(x)$ behave for values of $x$ near $-\frac{\pi}{2}$?
3. What is the limit $\lim_{x \to \pi} \sec(x)$?
4. Can the secant function ever be negative? If so, when?
5. How is the secant function related to trigonometric identities?

Tip:

The secant function has vertical asymptotes where the cosine function equals zero, which occurs at $\frac{\pi}{2} + n\pi$, where $n$ is an integer.