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 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?

 

[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 \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?

<br /> <br /> 1- (\sqrt{x+1})^2= x[/itex]&lt;br /&gt; Ok, I don&amp;#039;t see this. isn&amp;#039;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 \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?

<br /> <br /> Limit = 0

 

[arildno]

1- (\sqrt{x+1})^2= x[/itex]<br /> Ok, I don&#039;t see this. isn&#039;t it just -x because 1-(x+1)=-x?

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

 

[kathrynag]

HallsofIvy just forgot the minus sign.

Ok, i thought that was just it.