Defining f(0) to be continuous

[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.

[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}}

[kathrynag]

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}}

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

[HallsofIvy]

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.

[kathrynag]

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.

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

[HallsofIvy]

Yes, of course!

[kathrynag]

Ok thanks!

[kathrynag]

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

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

[HallsofIvy]

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?

[arildno]

The numerator is -x, Halls, not x!

[kathrynag]

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?

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

[kathrynag]

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?

Limit = 0

[arildno]

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

HallsofIvy just forgot the minus sign.

[kathrynag]

HallsofIvy just forgot the minus sign.

Ok, i thought that was just it.