Need help proving a limit using epsilon delta definition.

[nickadams]

Need help proving lim(x)->(a) sqrt(x)=sqrt(a) using epsilon delta definition.

Homework Statement

Prove that the limit of \sqrt{x} is \sqrt{a} as x approaches a
if a>0

Homework Equations

in words

By the epsilon delta definition we know that the distance between f(x) and the limit of the function will be less than "epsilon" if x is within a certain distance "delta" of a.


in equation form

0<|x-a|<delta
|f(x)-L|<epsilon

The Attempt at a Solution

ok i got messed up with the latex and i don't know how to fix it.

But what I did was multiplied |sqrt(x)-sqrt(a)| by |sqrt(x)+sqrt(a)| over itself to get the top in a form similar to delta. So that comes out to |x-a|/|sqrt(x)+sqrt(a)| is less than epsilon.

so we have |x-a| in both the epsilon and delta inequalities, so is it safe to write delta in terms of epsilon? I don't know what to do now...
?




i'm stuck

 

[nickadams]

ok i got messed up with the latex and i don't know how to fix it.

But what I did was multiplied |sqrt(x)-sqrt(a)| by |sqrt(x)+sqrt(a)| over itself to get the top in a form similar to delta. So that comes out to |x-a|/|sqrt(x)+sqrt(a)| is less than epsilon.

so we have |x-a| in both the epsilon and delta inequalities, so is it safe to write delta in terms of epsilon? I don't know what to do now...

 

[nickadams]

Prove that the limit of √x as x approaches a is √a
if a>0


please help...

 

[SammyS]

Let's fix the Latex problem.

...

The Attempt at a Solution

0<|x-a|<\delta
|\sqrt{x}-\sqrt{a}|<\epsilon

using (a-b)(a+b)=(a^{2}-b^{2}) the above epsilon inequality is changed to...

\frac{|x-a|}{|\sqrt{x}+\sqrt{a}|}&lt;\epsilon

now |x-a| is present in both the \epsilon and \delta equations so...?

I fixed the following, too.

|\delta|<\epsilon\cdot|\sqrt{x}+\sqrt{a}|

?

i'm stuck

 

[nickadams]

Thank you so much SammyS!

I have made further progress since I last posted and I may be coming to a solution. I will post what I have soon and hopefully I can get the latex right this time.

 

[nickadams]

Ok, I got stuck again. Here is what I have...

0<|x-a|< delta
|\sqrt{x}-\sqrt{a}| < epsilon

multiply |\sqrt{x}-\sqrt{a}|
by
|\sqrt{x}+\sqrt{a}| over itself
to get
|x-a|/\sqrt{x}+\sqrt{a} < epsilon

Now we need to find a constant C such that
|x-a|/\sqrt{x}+\sqrt{a} < |x-a|/C < epsilon

Since we want |x-a|, (AKA "delta") to be a pretty small number, we will choose it to be <1

in order for |x-a|/C to be > than |x-a|/\sqrt{x}+\sqrt{a}, C must be smaller than |\sqrt{x}+\sqrt{a}|

for the next step we will set up the inequality
|x-a|< 1
and try to rearrange it in a way where we can discover what the constant C (which is < than |\sqrt{x}+\sqrt{a}|) could be set equal to to assist in solving the problem.

solving |x-a|< 1 in terms of x gives us
x < a\pm1

so since we can set up this inequality
C < |\sqrt{x}+\sqrt{a}| < |sqrt{a\pm1}+sqrt{a}

but now i am stuck. Where have I gone wrong?

 

[HallsofIvy]

You want |\sqrt{x}- \sqrt{a}|&lt;\epsilon. Do NOT "multiply by \sqrt{x}+ \sqrt{a} over itself". Instead, multiply both sides by \sqrt{x}+ \sqrt{a} (which is obviously positive) to get |x- a|&lt; \epsilon(\sqrt{x}+ \sqrt{a}). Since \sqrt{x} is positive, \sqrt{a}&lt; \sqrt{x}+ \sqrt{a} and \epsilon\sqrt{a}&lt; \epsilon(\sqrt{x}+ \sqrt{a}). You can take \delta= \epsilon\sqrt{a}. That way, if |x- a|&lt; \delta= \epsilon\sqrt{a} you must also have |x- a|&lt; \epsilon(\sqrt{x}+\sqrt{a}).

 

[nickadams]

You want |\sqrt{x}- \sqrt{a}|&lt;\epsilon. Do NOT "multiply by \sqrt{x}+ \sqrt{a} over itself". Instead, multiply both sides by \sqrt{x}+ \sqrt{a} (which is obviously positive) to get |x- a|&lt; \epsilon(\sqrt{x}+ \sqrt{a}). Since \sqrt{x} is positive, \sqrt{a}&lt; \sqrt{x}+ \sqrt{a} and \epsilon\sqrt{a}&lt; \epsilon(\sqrt{x}+ \sqrt{a}). You can take \delta= \epsilon\sqrt{a}. That way, if |x- a|&lt; \delta= \epsilon\sqrt{a} you must also have |x- a|&lt; \epsilon(\sqrt{x}+\sqrt{a}).

Thanks HallsofIvy. Although I can see why every one of your statements is true, I don't understand how it proves the limit as x approaches a of \sqrt{x} is \sqrt{a}.

I just don't understand why you did any of the things that you did. :(
Could you (or someone else) maybe help me understand what we were trying to accomplish? I know the problem wants us to prove a limit using the epsilon delta definition of a limit, but I don't understand what that entails, and I certainly wouldn't have known what steps to take if you hadn't shown me...




Edit: Are we able to take \delta= \epsilon\sqrt{a} because we know that |x- a|&lt; \delta and |x- a|&lt; \epsilon(\sqrt{x}+ \sqrt{a}), and \epsilon\sqrt{a}&lt; \epsilon(\sqrt{x}+ \sqrt{a}), so since |x-a| is less than \delta and |x-a| and \epsilon\sqrt{a} are both less than \epsilon(\sqrt{x}+ \sqrt{a}), we are able to set \delta equal to \epsilon\sqrt{a}? But if \delta = \epsilon\sqrt{a}, wouldn't that mean that \epsilon\sqrt{a} > |x-a|? And how can we know that?

 

[HallsofIvy]

To prove that "\lim_{x\to a}f(x)= L" means to show that, given any \epsilon&gt; 0 there exist \delta&gt; 0 such that if |x- a|&lt;\delta then |f(x)- L|&lt; \epsilon.

The point of choosing \delta= \epsilon\sqrt{a} is that you can then work backwards from the way I showed before:
If |x- a|&lt; \epsilon\sqrt{a}&lt; \epsilon (\sqrt{x}+\sqrt{a})
then
\frac{|x- a|}{\sqrt{x}+ \sqrt{a}}&lt; \epsilon
\frac{|(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})|}{\sqrt{x}+ \sqrt{a}}&lt; \epsilon
|\sqrt{x}- \sqrt{a}|&lt; \epsilon

 

[nickadams]

To prove that "\lim_{x\to a}f(x)= L" means to show that, given any \epsilon&gt; 0 there exist \delta&gt; 0 such that if |x- a|&lt;\delta then |f(x)- L|&lt; \epsilon.

The point of choosing \delta= \epsilon\sqrt{a} is that you can then work backwards from the way I showed before:
If |x- a|&lt; \epsilon\sqrt{a}&lt; \epsilon (\sqrt{x}+\sqrt{a})
then
\frac{|x- a|}{\sqrt{x}+ \sqrt{a}}&lt; \epsilon
\frac{|(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})|}{\sqrt{x}+ \sqrt{a}}&lt; \epsilon
|\sqrt{x}- \sqrt{a}|&lt; \epsilon

Okay that makes sense too. But where do we go from there to prove that the limit of √x is √a as x approaches a if a>0?

Can I say: If we want to be within |\sqrt{x}- \sqrt{a}| of the limit of √x, then we must pick an x value that is within |x-a| of a (AKA delta). So the delta we choose will be equal to \epsilon *|\sqrt{x}- \sqrt{a}|...

now what?

 

[SammyS]

Okay that makes sense too. But where do we go from there to prove that the limit of √x is √a as x approaches a if a>0?

Can I say: If we want to be within |\sqrt{x}- \sqrt{a}| of the limit of √x, then we must pick an x value that is within |x-a| of a (AKA delta). So the delta we choose will be equal to \epsilon *|\sqrt{x}- \sqrt{a}|...

now what?

What HallsofIvy gave you in post #7 is what is often done on "scratch-paper", working backwards from |f(x) - L| < ε to |x-a| < δ_ε , where δ_ε is some expression for δ which is (usually) based on ε .

In final version of the proof you reverse that backwards argument and clean up any rough spots.

In other words: Using the δ_ε that you found, start with 0 < |x - a| < δ_ε. Then reversing your backwards "scratch-paper" argument, eventually you should arrive at |f(x) - L| < ε .

In your case, HallsofIvy showed you that δ_ε = ε∙√a should give you your desired result.​