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

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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!

User KishanCS
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