Defining f(0) to be continuous

[kathrynag]

Homework Statement

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?

Homework Equations

lx-x₀l<delta
lf(x)-f(x₀l<epsilon

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 [tex]1- (\sqrt{x+1})^2= x[/itex] and the denominator is $$\sqrt{x}(1+ \sqrt{x+1})$$. Now, $$x/\sqrt{x}$$ is $$\sqrt{x}$$ so the fraction becomes <br /> $$\frac{\sqrt{x}}{1+ \sqrt{x+1}}$$<br /> What is the limit of that as x goes to 0?[/tex]

 

[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 [tex]1- (\sqrt{x+1})^2= x[/itex] and the denominator is $$\sqrt{x}(1+ \sqrt{x+1})$$. Now, $$x/\sqrt{x}$$ is $$\sqrt{x}$$ so the fraction becomes <br /> $$\frac{\sqrt{x}}{1+ \sqrt{x+1}}$$<br /> What is the limit of that as x goes to 0?[/tex]

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

 

[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 [tex]1- (\sqrt{x+1})^2= x[/itex] and the denominator is $$\sqrt{x}(1+ \sqrt{x+1})$$. Now, $$x/\sqrt{x}$$ is $$\sqrt{x}$$ so the fraction becomes <br /> $$\frac{\sqrt{x}}{1+ \sqrt{x+1}}$$<br /> What is the limit of that as x goes to 0?[/tex]

$$<br /> Limit = 0$$

 

[arildno]

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

$$<br /> HallsofIvy just forgot the minus sign.$$

 

[kathrynag]

HallsofIvy just forgot the minus sign.

Ok, i thought that was just it.