Real Analysis: Proof about simple zeros (multiplicity = 1)

[GangSines]

Homework Statement

A function f has a simple zero (or zero of multiplicity 1) at x₀ if f is differentiable in a neighborhood of x₀ and f(x₀) = 0 while f(x₀) ≠ 0.

Prove that f has a simple zero at x₀ iff f(x) = g(x)(x - x₀), where g is continuous at x₀ and differentiable in a deleted neighborhood of x₀, and g(x₀) ≠ 0.

Homework Equations

None.

The Attempt at a Solution

You need to find a function g(x) such that it fufills the properties in the problem. I thought of using the function g(x) = f(x)/(x-x₀), but that is not continuous at x₀. Am I close in trying to find this function g or should the proof go in a different direction?

 

[gopher_p]

You're on the right track. Can you show that $\lim_{x\rightarrow x_0}\frac{f(x)}{x-x_0}$ exists?

 

[GangSines]

After using L'hospital's rule, you find that the limit is f'(x₀) ≠ g(x₀). So g is not continuous at x₀.

 

[gopher_p]

After using L'hospital's rule, you find that the limit is f'(x₀) ≠ g(x₀). So g is not continuous at x₀.

How is g(x₀) defined?

 

[GangSines]

My idea was to have g(x) = f(x)/(x-x0), that satisfies f(x) = g(x)(x - x0), but not the continuity part

 

[gopher_p]

My idea was to have g(x) = f(x)/(x-x0), that satisfies f(x) = g(x)(x - x0), but not the continuity part

That formula, f(x)/(x-x₀), defines g(x) for x≠x₀ and is undefined at x₀. How are you defining g(x₀)? What is preventing you from making g(x₀)=f'(x₀) (or any other number, for that matter)?

 

[GangSines]

I would just have to choose that number such that it makes g(x0) continuous?

 

[gopher_p]

I would just have to choose that number such that it makes g(x0) continuous?

Right. And that number would be ...

Also (and I apologize for the pedantry), you should say (and write) "g continuous at x₀" rather than "g(x₀) continuous". g(x₀) is a number, and continuity is not a property of numbers.