# Defining f(0) to be continuous

1. Nov 15, 2008

### kathrynag

1. The problem statement, all variables and given/known data
f(0,1) ---> R by f(x) =1/x^(1/2) -((x+1)/x)^(1/2). Can one define f(0) to make f continuous at 0?

2. Relevant equations
lx-x0l<delta
lf(x)-f(x0l<epsilon

3. The attempt at a solution
My thought is that the limit must equal f(0), but I'm unsure of how to get f(0) because of division by zero.

2. Nov 15, 2008

### arildno

Write this as a single fraction:
$$f(x)=\frac{1-\sqrt{x+1}}{\sqrt{x}}$$
Now, utilize the following identity in a constructive fashion:
$$1=\frac{1+\sqrt{x+1}}{1+\sqrt{x+1}}$$

3. Nov 15, 2008

### kathrynag

ok, so $$\frac{x-x^{2}}{\sqrt{x+1}}$$

4. Nov 15, 2008

### HallsofIvy

Staff Emeritus
And? Since f(x) and that are exactly the same for all x except x= 0, they have the same limit at x=0. What is the limit as x goes to 0? Define f(0) to be that limit.

5. Nov 15, 2008

### kathrynag

Ok, so the limit = 0. Let f(0)=0?

6. Nov 15, 2008

### HallsofIvy

Staff Emeritus
Yes, of course!

7. Nov 15, 2008

### kathrynag

Ok thanks!

8. Nov 16, 2008

### kathrynag

Ok, I think I did this wrong as I look at it today. Can anybody help me simplify a bit?

9. Nov 16, 2008

### HallsofIvy

Staff Emeritus
You are multiplying numerator and denominator of
$$f(x)=\frac{1-\sqrt{x+1}}{\sqrt{x}}$$
by $$1+ \sqrt{x+1}$$ so then numerator becomes $$1- (\sqrt{x+1})^2= x[/itex] and the denominator is [tex]\sqrt{x}(1+ \sqrt{x+1})$$. Now, $$x/\sqrt{x}$$ is $$\sqrt{x}$$ so the fraction becomes
$$\frac{\sqrt{x}}{1+ \sqrt{x+1}}$$
What is the limit of that as x goes to 0?

10. Nov 16, 2008

### arildno

The numerator is -x, Halls, not x!

11. Nov 16, 2008

### kathrynag

[tex]1- (\sqrt{x+1})^2= x[/itex]
Ok, I don't see this. isn't it just -x because 1-(x+1)=-x?

12. Nov 16, 2008

### kathrynag

Limit = 0

13. Nov 16, 2008

### arildno

HallsofIvy just forgot the minus sign.

14. Nov 16, 2008

### kathrynag

Ok, i thought that was just it.