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Let \(\mathrm{f}: \mathrm{R} \rightarrow \mathrm{R}\) be given by \(\mathrm{f}(\mathrm{x})=\mathrm{x}^{\mathrm{n}}\) where $\mathrm{n} \in \mathrm{N}$, the set of natural numbers. Prove that \(\mathrm{f}\) is everywhere continuous.

Short Answer

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We proved that the function \(f(x) = x^n\) is continuous for all x in ℝ and for all natural number values of n by using the ε-δ definition of continuity. We factored the expression \(|f(x) - f(a)| = |x^n - a^n|\) and found a bound M on the factored expression. Then, we chose δ = min(1, ε / (n * M)) to ensure that if \(|x - a| < \delta\), then \(|f(x) - f(a)| < \varepsilon\). Since this holds for all ε > 0, the function is continuous everywhere.
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Step 1: Recall the ε-δ definition of continuity

A function f is continuous at a point x = a if for every ε > 0, there exists a δ > 0 such that, if |x-a| < δ, then |f(x)-f(a)| < ε. Our goal is to prove that f(x) = x^n is continuous for all x in ℝ and for all natural number values of n.

Step 2: Show that |f(x) - f(a)| can be factored for f(x) = x^n

In order to use the ε-δ definition of continuity, we need to find a way to express |f(x) - f(a)| in relation to |x - a|. Let f(x) = x^n. Then f(a) = a^n So, |f(x) - f(a)| = |x^n - a^n| Factor the expression as follows: |x^n - a^n| = |(x - a)(x^(n-1) + a*x^(n-2) + ... + a^(n-1))|

Step 3: Find a bound for the factored expression

Next, we want to find a bound on the expression |x^(n-1) + a*x^(n-2) + ... + a^(n-1)| to relate it to |x - a|. Notice that each term in the expression is less than or equal to M = max(|x^(n-1)|, |a^(n-1)|). Thus, we have: |x^(n-1) + a*x^(n-2) + ... + a^(n-1)| ≤ n * M |x^n - a^n| = |(x - a)(x^(n-1) + a*x^(n-2) + ... + a^(n-1))| ≤ |x - a| * n * M

Step 4: Apply the ε-δ definition of continuity

Now let ε > 0 be given. We want to find δ > 0 such that, if |x - a| < δ, then |f(x) - f(a)| < ε. Using the bound we found in step 3, we can say that if |x - a| < δ, then |f(x) - f(a)| ≤ |x - a| * n * M. To ensure that |f(x) - f(a)| < ε, we need to choose δ such that if |x - a| < δ, then |x - a| * n * M < ε. Let δ = min(1, ε / (n * M)) Then |x - a| < δ implies |x - a| * n * M < ε, thus |f(x) - f(a)| < ε.

Step 5: Conclusion

We have shown that, for every ε > 0, there exists a δ > 0 such that if |x - a| < δ, then |f(x) - f(a)| < ε. This satisfies the ε-δ definition of continuity, and we can conclude that the function f(x) = x^n is continuous for all x in ℝ and for all natural number values of n.

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