Stuck on proof- show |f(x)-f(x0)|<|x-x0| if f(x)=sqrt(a+x^2) and x≠x0

[natasha13100]

Homework Statement

Hey. I am taking my first Real Analysis course at the undergraduate level and am not really sure where to start on a problem. I do have some ideas. I am not asking anyone to solve it for me, but I do need an idea where to start.

Show that |f(x)-f(x₀)|<|x-x₀| if f(x)=√(a+x²) where a is a constant and x≠x₀. What does this prove about x?

Homework Equations

I'm not sure.
We've talked about limits, inequalities, induction, reals, nested intervals, rational and irrational numbers, upper and lower bounds.
More recently we discussed continuity, boundedness, extreme values, and intermediate value theorem.

The Attempt at a Solution

Here are some ideas that came to mind when I tried to start the problem:
1. Given ε>0, there is 0<|x-x₀|<δ so that |f(x)-A|<ε where A is the limit as x→x₀ of f(x) if f is defined at x.
2. If x>x₀, f(x)=√(a+x²)>√(a+x₀²)=f(x₀)→x²>x₀².
So x²-x₀²>0 and x-x₀>0.
If x<x₀, f(x)=√(a+x²)<√(a+x₀²=f(x₀)→x²<x₀².
So x²-x₀²<0 and x-x₀<0.

I don't even think these are correct. Where can I start?

 

[Mark44]

Homework Statement

Hey. I am taking my first Real Analysis course at the undergraduate level and am not really sure where to start on a problem. I do have some ideas. I am not asking anyone to solve it for me, but I do need an idea where to start.

Show that |f(x)-f(x₀)|<|x-x₀| if f(x)=√(a+x²) where a is a constant and x≠x₀. What does this prove about x?

Homework Equations

I'm not sure.
We've talked about limits, inequalities, induction, reals, nested intervals, rational and irrational numbers, upper and lower bounds.
More recently we discussed continuity, boundedness, extreme values, and intermediate value theorem.

The Attempt at a Solution

Here are some ideas that came to mind when I tried to start the problem:
1. Given ε>0, there is 0<|x-x₀|<δ so that |f(x)-A|<ε where A is the limit as x→x₀ of f(x) if f is defined at x.
2. If x>x₀, f(x)=√(a+x²)>√(a+x₀²)=f(x₀)→x²>x₀².
So x²-x₀²>0 and x-x₀>0.
If x<x₀, f(x)=√(a+x²)<√(a+x₀²=f(x₀)→x²<x₀².
So x²-x₀²<0 and x-x₀<0.

I don't even think these are correct. Where can I start?

Are you sure you copied this part correctly?

What does this prove about x?

I would think that the question would be asking about f(x), not x.

For the given function, the statement |f(x)-f(x₀)|<|x-x₀|, x≠x₀ is equivalent to saying

$$\frac{|f(x) - f(x_0)|}{|x - x_0|} < 1$$

I would replace f(x) and f(x₀ with the function you're given and see if I could show that the numerator is always less than the denominator, which would make the quotient less than 1.