Question

If 3x ≤ f(x) ≤ x^3 + 2 for 0 ≤ x ≤ 2,, Find Lim x →1f(x).

Answer 1

Given inequality:

`\sf\:3x \leq f(x) \leq x^3 + 2 \quad \text{for } 0 \leq x \leq 2 \\`

To find the limit as x approaches 1 of f(x), we can use the Squeeze Theorem. Since
`\sf\:3x \leq f(x) \leq x^3 + 2 \\` holds for
`\sf\:0 \leq x \leq 2 \\`, we can evaluate the limits of the lower and upper bounds and check if they are equal at x = 1.

1. Lower bound: 3x

`\sf\:\lim_{{x \to 1}} 3x = 3 \cdot 1 = 3 \\`

2. Upper bound:
`\sf\:x^3 + 2 \\`

`\sf\:\lim_{{x \to 1}} (x^3 + 2) = (1^3 + 2) = 3 \\`

Since the limits of both the lower and upper bounds are equal to 3 at x = 1, we can conclude that:

`\sf\:\lim_{{x \to 1}} f(x) = 3 \\`

That's it!