%3Cp%3EProve%20using%26nbsp%3BEpsilon%26nbsp%3Bdelta%26nbsp%3BDefinitions%3C%2Fp

Question

%3Cp%3EProve%20using%26nbsp%3BEpsilon%26nbsp%3Bdelta%26nbsp%3BDefinitions%3C%2Fp%3E%3Cp%3E%3Cbr%3E%3C%2Fp%3E%3Cp%3E%3Cimg%26nbsp%3Bundefined%3D%22%22%26nbsp%3Bclass%3D%22user-upload%22%26nbsp%3Bsrc%3D%22http%3A%2F%2Fmedia.cheggcdn.com%2Fmedia%252F512%252F512908a1-3cd0-4605-a16e-53785b8318b2%252Fphp4l7z5o.png%22%26nbsp%3Bdata-mce-src%3D%22http%3A%2F%2Fmedia.cheggcdn.com%2Fmedia%252F512%252F512908a1-3cd0-4605-a16e-53785b8318b2%252Fphp4l7z5o.png%22%26nbsp%3Bheight%3D%22width%3D%22%26nbsp%3Bwidth%3D%22undefined%22%3E%3C%2Fp%3E

Explanation / Answer

f(x)=sqrt(x)-x for

Let a>0 and take x in R+

|f(x)-f(a)|= |sqrt(x)-sqrt(a)+ a-x| <= |sqrt(x)-sqrt(a)|+|x-a| by triangular inequality

now notice that |x-a| = | sqrt(x)-sqrt(a)||sqrt(x)+sqrt(a)| so :

|sqrt(x)-sqrt(a)| = |x-a|/(sqrt(x)+sqrt(a)) <= |x-a|/sqrt(a) since x>=0 and a>0

So |f(x)-f(a)| <= |x-a| (1 + 1/sqrt(a))

Now let epsilon > 0 and set delta= epsilon/(1+1/sqrt(a))

Then |x-a|< delta => |f(x)-f(a)| <= delta * (1+1/sqrt(a)) = epsilon.

So f is continuous at every a>0.

Related Questions

Navigate

• Browse (All)
• Browse #
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.