Use the definition of continuity to prove that the function f defined by f(x) = (x)^(½) is continuous at every nonnegative real number.
Our definition of continuity: Let I be an interval, let f:I-R, and let c(element of)I. The function f is continuous at c if for each (epsilon)>0 there exists (delta)>0 such that |f(x)-f(c)| < (epsilon) for all x (element of) I that satisfy |x-c|<(delta). The function f is continuous on I if f is continuous at each point of I.
I'm not sure how to the nonnegative real number comes into play here in terms of proving the continuity.
Best Answer
Continuity at $c=0$ is continuity from the right, so one wants to show given $\varepsilon$ there is $\delta$ for which $0 \le x < \delta$ implies $0 \le \sqrt{x} < \varepsilon$. The natural choice here is $\delta=\varepsilon^2.$
Otherwise $c>0$ and we initially restrict to $\delta<c/2$ (and throw that in at the end via the usual "min" definition.) From this assumption we have $c/2<x<3c/2$ and so $$\sqrt{c/2}<\sqrt{x}<\sqrt{3c/2}.$$
Now we consider that $$|\sqrt{x}-\sqrt{c}|=\frac{|x-c|}{\sqrt{x}+\sqrt{c}}. \tag{1}$$ Given our bounds for $\sqrt{x}$ the right side of $(1)$ is at most $|x-c|/(\sqrt{c/2}+\sqrt{c}),$ so that if we finally define $\delta$ as the minimum of the two numbers $c/2$ and $\varepsilon \cdot (\sqrt{c/2}+\sqrt{c})$ we'll have $|x-c|<\delta$ implies $|\sqrt{x}-\sqrt{c}|<\varepsilon.$