prove by epsilon-delta definition that f:R->R given by f(x)=x^3 is continuous at x=2

Expert Solution

Proof : f : R R is continuous at x₀ if for all > 0 there exists > 0 for all x R:

Therefore f(x) = x³ and x₀ = 2

Therefore for all > 0 there exists > 0 for all x R > 0

( | x - 2 | < | x³ - 2³ | = | (x-2)(x² + 2x + 4)| < where is bigger

| (x-2)(x² + 2x + 4)| =

Therefore f(x) = x³ is continuous at x = 2

Colby Messinger answered 2 years ago

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