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...

$\frac{|x-a|}{|\sqrt{x}+\sqrt{a}|}<\epsilon$

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$\pm$1

so since we can set up this inequality
C < |$\sqrt{x}$+$\sqrt{a}$| < |sqrt{a$\pm$1}+sqrt{a}

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

 

[HallsofIvy]

You want $|\sqrt{x}- \sqrt{a}|<\epsilon$. 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 $|x- a|< \epsilon(\sqrt{x}+ \sqrt{a})$. Since $\sqrt{x}$ is positive, $\sqrt{a}< \sqrt{x}+ \sqrt{a}$ and $\epsilon\sqrt{a}< \epsilon(\sqrt{x}+ \sqrt{a})$. You can take $\delta= \epsilon\sqrt{a}$. That way, if $|x- a|< \delta= \epsilon\sqrt{a}$ you must also have $|x- a|< \epsilon(\sqrt{x}+\sqrt{a})$.

 

[nickadams]

You want $|\sqrt{x}- \sqrt{a}|<\epsilon$. 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 $|x- a|< \epsilon(\sqrt{x}+ \sqrt{a})$. Since $\sqrt{x}$ is positive, $\sqrt{a}< \sqrt{x}+ \sqrt{a}$ and $\epsilon\sqrt{a}< \epsilon(\sqrt{x}+ \sqrt{a})$. You can take $\delta= \epsilon\sqrt{a}$. That way, if $|x- a|< \delta= \epsilon\sqrt{a}$ you must also have $|x- a|< \epsilon(\sqrt{x}+\sqrt{a})$.

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 $|x- a|< \delta$ and $|x- a|< \epsilon(\sqrt{x}+ \sqrt{a})$, and $\epsilon\sqrt{a}< \epsilon(\sqrt{x}+ \sqrt{a})$, 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 $\epsilon> 0$ there exist $\delta> 0$ such that if $|x- a|<\delta$ then $|f(x)- L|< \epsilon$.

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

 

[nickadams]

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

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

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.​