Question

Define f:R→R by f(x)=x² . Show that f is continuous at 1 . For any given (particular but arbitrarily chosen) ε∈R + , let δ= min{1,ε/3}∈R+ . Then, for any x∈D=R, if ∣x−1∣<δ=min{1,ε/3}

Answer 1

The function f(x) = x² is proven to be continuous at x = 1 using the ε-δ definition of continuity by showing that for every ε > 0 there exists a δ > 0 such that if |x - 1| < δ, then |f(x) - f(1)| < ε.

Step-by-step explanation:

To demonstrate that the function f(x) = x² is continuous at x = 1, we utilize the ε-δ definition of continuity. For any given ε > 0, we need to find a δ > 0 such that if |x - 1| < δ, then |f(x) - f(1)| < ε. We choose δ = min{1, ε/3}. This ensures that for any x within δ of 1, the change in f(x) is within ε of f(1), thus confirming the continuity at x = 1.

If |x - 1| < δ, then |x - 1| < ε3, and we can find the bound for |f(x) - f(1)|:

• |f(x) - f(1)| = |x² - 1²| = |(x - 1)(x + 1)|
• ≤ |x - 1||x + 1|
• Since |x - 1| < 1, we have -1 < x < 2, so |x + 1| ≤ 3.
• Therefore, |f(x) - f(1)| ≤ 3|x - 1| < 3(ε/3) = ε.

Thus, the function f(x) = x² is continuous at x = 1, as for every ε > 0, a δ can be found such that the above conditions hold.