How to Prove a Limit in Calculus

[djh101]

Homework Statement

Prove the limit. I'm not entirely sure how to prove limits, the book doesn't go into much detail.

Homework Equations

Lim x→9 √(x-5) = 2

The Attempt at a Solution

Prove |x - 9| < δ

|√(x-5) - 2| < ε
2 - ε < √(x - 5) < 2 + ε
ε² - 4ε + 4 < x - 5 < ε² + 4ε + 4
ε² - 4ε < x - 9 < ε² + 4ε
δ = ε² + 4ε

 

[tylerc1991]

Lim x→9 √(x-5) = 2

For proving limits, you need to show that, for all $\varepsilon > 0$, there exists a $\delta > 0$ such that

$|x - 9| < \delta \implies |\sqrt{x - 5} - 2| < \varepsilon$

So to begin, choose an $\varepsilon > 0$. Then you need to choose a $\delta > 0$ such that the above implication is true.

 

[LCKurtz]

Homework Statement

Prove the limit. I'm not entirely sure how to prove limits, the book doesn't go into much detail.

Homework Equations

Lim x→9 √(x-5) = 2

The Attempt at a Solution

Prove |x - 9| < δ

No, that isn't what you are trying to prove. The problem is to figure out a δ > 0 that will make your next inequality true if 0 < |x - 9| < δ.

|√(x-5) - 2| < ε

You can tell by looking that if x is close to 9 then |√(x-5) - 2| will be close to zero, but how close x needs to be to 9 depends on how small ε is. The next few steps might be called an exploratory argument

2 - ε < √(x - 5) < 2 + ε
ε² - 4ε + 4 < x - 5 < ε² + 4ε + 4
ε² - 4ε < x - 9 < ε² + 4ε
δ = ε² + 4ε

Without multiplying those squares out you have

(2-ε)² - 4 < x - 9 < (2+ε)² - 4

Notice that the left side is negative (at least if ε < 2) and the right side is positive, so this is almost n the form

-δ < x - 9 < δ

but the δ on the left would be 4 - (2-ε)² and the δ on the right is (2+ε)² - 4

Figure out which of those is the smaller and use it for δ. Then you can reverse your argument like this:

Given ε > 0 let δ = [your minimum value here]. Then if |x - 9| < δ [work your steps backwards here to end up with |√(x-5) - 2| < ε.

 

[djh101]

Okay, so I've come down to δ₁ = -ε² + 4ε and δ₂ = ε². δ₁ will be smaller so δ = -ε² + 4ε.

So, $\forall$ ε > 0, $\exists$ δ = -ε² + 4ε > 0 such that |$\sqrt{x - 5}$ - 2| < ε whenever |x - 9| < δ, therefore Lim x→4 (9-x) = 5.

Am I on the right track?

 

[LCKurtz]

Okay, so I've come down to δ₁ = -ε² + 4ε and δ₂ = ε². δ₁ will be smaller so δ = -ε² + 4ε.

So, $\forall$ ε > 0, $\exists$ δ = -ε² + 4ε > 0 such that |$\sqrt{x - 5}$ - 2| < ε whenever |x - 9| < δ, therefore Lim x→4 (9-x) = 5.

Am I on the right track?

Word it like this. Given ε > 0, let δ = -ε² + 4ε. (People will think WOW! Where did that come from because you aren't going to show them that scratch paper with the exploratory argument.) But at this point, you can't just assert |$\sqrt{x - 5}$ - 2| < ε whenever |x - 9| < δ because why would anyone believe you? You have to show the steps working backwards in your exploratory argument. So the next step would be to show, explaining how you know, that if |x - 9| < δ then

(2-ε)² - 4 < x - 9 < (2+ε)² - 4

so you can proceed working backwards to your conclusion:
|$\sqrt{x - 5}$ - 2| < ε

 

[djh101]

Okay, I think I get it. Thanks you.