3-xsquare<=g(x)<=3secx for all x find lim xtends to0g(x)

To find the limit of g(x) as x approaches 0, we need to analyze the given inequality:

3 - x^2 ≤ g(x) ≤ 3sec(x)

First, let's focus on the left-hand side of the inequality:

3 - x^2 ≤ g(x)

To determine the maximum possible value of g(x), we need to minimize the left-hand side of the inequality. The expression x^2 reaches its minimum value when x = 0, so we have:

3 - 0^2 ≤ g(x)

Simplifying further, we have:

3 ≤ g(x)

Now, let's consider the right-hand side of the inequality:

g(x) ≤ 3sec(x)

To find the minimum possible value of g(x), we need to maximize the right-hand side of the inequality. The sec(x) function achieves its maximum value when x = 0, which is equal to 1. Therefore, we have:

g(x) ≤ 3(1)

Simplifying:

g(x) ≤ 3

Combining the two inequalities, we find that:

3 ≤ g(x) ≤ 3

Since the lower bound of g(x) is equal to the upper bound, we can conclude that the limit of g(x) as x approaches 0 is 3.