U: Prove Limit $\lim_{x{\rightarrow}c}\sqrt{x}=\sqrt{c}$

[kreil]

Show \lim_{x{\rightarrow}c}\sqrt{x}=\sqrt{c}

In other words, \forall \epsilon > 0 \exists \delta > 0 s.t. |x-c|< \delta \implies |\sqrt{x}-\sqrt{c}|< \epsilon

So 0<|x-c|<\delta \implies |\sqrt{x}-\sqrt{c}|=|\frac{x-c}{\sqrt{x}+\sqrt{c}}|

and this is where I am stuck, a hint the teacher gave was "bound the denominator" but I am not sure how to do that...can anyone help?

Josh

 

[StatusX]

You want to show:

\frac{|x-c|}{\sqrt{x}+\sqrt{c} } < \epsilon

The way these things work is that you want to find something bigger than the LHS that is still smaller than RHS, because then the LHS is definitely smaller than the RHS. So keep replacing the LHS by something bigger and bigger until it's in a form that is easy to compare to the RHS. What can you replace the denominator by (that's independent of x) to get something larger than the LHS?

 

[kreil]

\frac{|x-c|}{\sqrt{x}+\sqrt{c}}<\frac{|x-c|}{1}<\delta=\epsilon

Can I do that and just choose delta = epsilon?

 

[StatusX]

No, that isn't true. If x=c=1/16, then the denominator is 1/2, and so the LHS is not less than the middle. The bound will depend on c.

 

[kreil]

\frac{|x-c|}{\sqrt{x}+\sqrt{c}}<\frac{|x-c|}{3\sqrt{c}}

then this should work since if x=c=1/16, the LHS is 1/2 and the RHS is 3/4

can I then continue this so that

\frac{|x-c|}{3\sqrt{c}}<3\sqrt{c}\delta=\epsilon ...??

 

[StatusX]

No. First of all, you can't show something works with a single example. But it doesn't even work for that one. Remember, if you want to increase a fraction, you have to decrease its denominator. So you have to find something that is less than \sqrt{x}+\sqrt{c}.

 

[kreil]

so how about just squareroot of c?

 

[StatusX]

Yea, that's less than \sqrt{x}+\sqrt{c}. So what's your answer?

 

[kreil]

In other words, choose \delta=\frac{\epsilon}{\sqrt{c}}

\frac{|x-c|}{\sqrt{x}+\sqrt{c}}<\frac{|x-c|}{\sqrt{c}}<\sqrt{c} \delta = \epsilon

 

[StatusX]

Looks good.

 

[kreil]

Another limit question

I need to show that \lim_{x{\rightarrow}a}x^3=a^3

then 0<|x-a|<\delta \implies |x^3-a^3|< \epsilon

I'm having a lot of trouble getting the cubed terms to be less than an expression with |x-a|. Should I try turning the cubed terms into a fraction? I need a direction to go in.

Josh

 

[Muzza]

x^3 - a^3 = (x - a)(x^2 + ax + a^2).

 

[0rthodontist]

Also you could use |x|, |a| < |a| + e. That turns the second factor into something less than 3(a + e)^2

 

[kreil]

Yet Another E-d proof

I got that one now, thanks.

Now this one is giving me a problem...

Show \lim_{x{\rightarrow}1}\frac{x^2-x+1}{x+1}=\frac{1}{2}

i.e. show \forall \epsilon &gt; 0 \exists \delta &gt; 0 s.t. |x-1|&lt; \delta \implies |\frac{x^2-x+1}{x+1}-\frac{1}{2}|&lt;\epsilon

The teacher says to let x=1+ \lambda so that as x \rightarrow 1, \lambda \rightarrow 0. I wasn't quite sure how to properly incorporate this into the proof, because it gave me problems choosing delta:

|\frac{x^2-x+1}{x+1}-\frac{1}{2}|=|\frac{2x^2-3x+1}{2(x+1)}<br /> |=|\frac{(1+ \lambda -1)(2+2 \lambda -1)}{2(1+ \lambda +1)}|=|\frac{(\lambda)(2 \lambda +1)}{2( \lambda + 2)}=0&lt; \epsilon

I know that this can't be right since it doesn't even require me to choose a delta, so it doesn't really satisfy the definition. Any help incorporating the hint will be helpful.

 

[0rthodontist]

lambda is not equal to 0, lambda is some constant. You want to show that when 0 < |lambda| < delta for some delta, your expression on the right is less than epsilon.