The limit you're asking for is:

$$ \lim_{x \to 0} \sec(x) $$

First, recall the definition of the secant function:

$$ \sec(x) = \frac{1}{\cos(x)} $$

As $ x \to 0 $, we know that:

$$ \cos(0) = 1 $$

Thus:

$$ \sec(0) = \frac{1}{\cos(0)} = \frac{1}{1} = 1 $$

Therefore, the limit is:

$$ \lim_{x \to 0} \sec(x) = 1 $$

Would you like further details or have any questions?

Here are 5 related questions for further exploration:
1. What is the limit of $ \sec(x) $ as $ x \to \pi/2 $?
2. How does $ \sec(x) $ behave near asymptotes?
3. What is the derivative of $ \sec(x) $?
4. How can we calculate $ \lim_{x \to 0} \csc(x) $?
5. What is the Taylor series expansion of $ \sec(x) $ around $ x = 0 $?

**Tip**: For trigonometric functions, it's always helpful to remember key values and behaviors at important points like 0, $ \pi/2 $, and $ \pi $.

Explore the limit of the secant function as x approaches 0, using the trigonometric identity for secant: $$ \sec(x) = \frac{1}{\cos(x)} $$. Learn key concepts of limits and trigonometric behavior, suitable for high school students in Grades 11-12.

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