# Thread: delta-epsilon proof, but it's not the proof that's stumped me!

1. ## delta-epsilon proof, but it's not the proof that's stumped me!

Hi all, first post. I'm starting university next fall as a physics/math double major and I am taking Calculus I this summer so that I can take Physics I in the fall.

Here's the problem, we've been doing delta-epsilon proofs of limits. The class just started last thursday. I know how to do these proofs generally, and went through the homework. However, the professor asked if there were any math majors, and like an idiot I raised my hand, therefore I was "expected" to do an extra homework problem that he left out because it was "too hard."

Here it is:

$\lim_{x \to \4} \frac{\sqrt{2x-1} }{ \sqrt{x-3} } = \sqrt{7}$

Construct delta-epsilon proof.

I feel kind of embarrassed because I feel like I should be able to do it..

Prework:

$0 < | x - 4 | < \delta => | \frac{\sqrt{2x-1} }{ \sqrt{x-3} } - \sqrt{7} | < \epsilon$

And....

What would you do with

$| \frac{\sqrt{2x-1} }{ \sqrt{x-3} } - \sqrt{7} |$

To get a | x - 4 | factor? As you can see, my problem is algebraic mainly...

Sorry if it's an easy one, I just don't know how to get my foot in the door!

Thanks!!

2. Originally Posted by OneMileCrash
Hi all, first post. I'm starting university next fall as a physics/math double major and I am taking Calculus I this summer so that I can take Physics I in the fall.

Here's the problem, we've been doing delta-epsilon proofs of limits. The class just started last thursday. I know how to do these proofs generally, and went through the homework. However, the professor asked if there were any math majors, and like an idiot I raised my hand, therefore I was "expected" to do an extra homework problem that he left out because it was "too hard."

Here it is:

$\lim_{x \to \4} \frac{\sqrt{2x-1} }{ \sqrt{x-3} } = \sqrt{7}$

Construct delta-epsilon proof.

I feel kind of embarrassed because I feel like I should be able to do it..

Prework:

0 < | x - 4 | $< \delta$=> | [tex]\frac{\sqrt{2x-1} }{ \sqrt{x-3} } - \sqrt{7}[/math | < $\epsilon$

And....

What would you do with

| $\frac{\sqrt{2x-1} }{ \sqrt{x-3} } - \sqrt{7}$|

To get a | x - 4 | factor? As you can see, my problem is algebraic mainly...

Sorry if it's an easy one, I just don't know how to get my foot in the door!

Thanks!!
Off-hand, I'd try two things.

First, add the square root of 7 to the fraction by getting a common denominator.

Second, rationalize the numerator by multiplying by its "reciprocal". See where that leads.

3. I'll give it a shot and report back. I've also fixed the BB Code formatting for the math, thanks!

4. Still not sure where I'm going..

| $\frac{\sqrt{2x+1} }{ \sqrt{x-3} } - \sqrt{7}}$|

So:

| $\sqrt{7} = \frac{\sqrt{7x-21}{ \sqrt{x-3} }$|

So:

| $\frac{\sqrt{2x+1} }{ \sqrt{x-3} } - \sqrt{7} }$|

=

| $\frac{\sqrt{2x-1} - \sqrt{7x-21} }{ \sqrt{x-3} }$|

=

| $\frac{\( \sqrt{x-3} )( \sqrt{2x-1}-\sqrt{7x-21}) }{ x-3}$|

Am I on the right track? Because I don't see this going anywhere.

5. Originally Posted by OneMileCrash
Still not sure where I'm going..

| $\frac{\sqrt{2x+1} }{ \sqrt{x-3} } - \sqrt{7}}$|

So:

| $\sqrt{7} = \frac{\sqrt{7x-21}{ \sqrt{x-3} }$|

So:

| $\frac{\sqrt{2x+1} }{ \sqrt{x-3} } - \sqrt{7} }$|

=

| $\frac{\sqrt{2x-1} - \sqrt{7x-21} }{ \sqrt{x-3} }$|

=

| $\frac{\( \sqrt{x-3} )( \sqrt{2x-1}-\sqrt{7x-21}) }{ x-3}$|

Am I on the right track? Because I don't see this going anywhere.
No, you need to rationalise the NUMERATOR, not the denominator.

So you should be multiplying top and bottom by $\displaystyle \sqrt{2x - 1} + \sqrt{7x - 21}$.

6. Curse my lack of attention-paying... but that leaves me with a ridiculous denominator, does it not?

$\frac{-5x-22}{(\sqrt{x-3} )(\sqrt{2x-1}+\sqrt{7x-21} )}$

The only other thing I could do is separate the last root into sqrt of 7 and sqrt of x-3 but I don't think that'll help.

I'm sorry guys, this algebra kills me.

7. I think you'll find that $\displaystyle (\sqrt{2x-1} - \sqrt{7x - 21})(\sqrt{2x - 1} + \sqrt{7x - 21}) = (2x - 1) - (7x - 21) = -5x + 20$

8. Right, silly me. And that is the first glimmer of something that is remotely similar to | x - 4 | when I pull the -5 out.

So now I'm left with

$\frac{-5(x+4)}{ (\sqrt{x-3} )(\sqrt{2x-1}+\sqrt{7x+21} )}$

If there is a way of simplifying that second parenthesis in the denominator, it's beyond me..as far as I can tell adding those 2 radicals is just a dead end. And it's a terrible feeling when you reach a dead end when you KNOW that there is some way to get to the solution. Can someone nudge me in regards to this denominator?

9. The numerator is $\displaystyle -5(x - 4)$...

10. For the record, THAT one is a typo. :/

11. since you only have to restrict your attention to x's that are within delta of 4, we can certainly require that delta be no greater than, say 1/2 (perhaps we will need it smaller).

in which case x-3 > 1/2, so √(x-3) > 1/√2 so 1/√(x-3) < √2 (see where i'm going with this?).

and since x > 3.5 > 3, 7x - 21 > 0 so √(2x-1) + √(7x - 21) > √(2x-1).

also since x > 3.5 > 1, 2x -1 > 1, so √(2x-1) > 1.

so $\frac{1}{(\sqrt{x-3})(\sqrt{2x-1} + \sqrt{7x-21})} < \sqrt{2}$ when $\delta < 1/2$

this is a fairly crude upper bound for this expression, but crude is all we need, because we can simultaneously impose another restriction on delta based on epsilon, and take the minimum.

12. We require finding a value of $\displaystyle M$ such that $\displaystyle \frac{1}{\sqrt{x-3}(\sqrt{2x-1} + \sqrt{7x+21})} \leq M$ for all possible $\displaystyle x$. To do this we have to restrict $\displaystyle x$, which we can do because $\displaystyle |x - 4|$ has to be made small anyway. Also note that since we are finding the maximum possible value for $\displaystyle M$, we are finding the minimum possible value for $\displaystyle \sqrt{x-3}(\sqrt{2x-1} + \sqrt{7x+21})$.

If we let $\displaystyle |x - 4| < \frac{1}{2}$ (this has to be made small anyway) then we find

Part 1:

\displaystyle \begin{align*} &\phantom{<} |x - 4|^2 < \frac{1}{4} \\ &\phantom{<} |(x - 4)^2| < \frac{1}{4} \\ -\frac{1}{4} &< (x - 4)^2 < \frac{1}{4} \\ -\frac{1}{4} &< x^2 - 8x + 16 < \frac{1}{4} \\ -\frac{1}{2} &< 2x^2 - 16x + 32 \\ 9x - \frac{59}{2} &< 2x^2 - 7x + 3 \end{align*}

and since we let $\displaystyle |x - 4| < \frac{1}{2} \implies -\frac{1}{2} < x - 4 \implies \frac{7}{2} < x \implies \frac{63}{2} < 9x \implies 2 < 9x - \frac{59}{2}$

that means

\displaystyle \begin{align*} 9x - \frac{59}{2} &< 2x^2 - 7x + 3 \\ 2 &< 2x^2 - 7x + 3 \\ \sqrt{2} &< \sqrt{2x^2 - 7x + 3}\end{align*}

Part 2: Since we let $\displaystyle |x - 4| < \frac{1}{2} \implies -\frac{1}{4} < (x - 4)^2$

\displaystyle \begin{align*} -\frac{1}{4} &< x^2 - 8x + 16 \\ -\frac{7}{4} &< 7x^2 - 56x + 112 \\ -\frac{707}{4} &< 7x^2 - 56x - 63 \\ 56x - \frac{707}{4} &< 7x^2 - 63 \end{align*}

and since we let $\displaystyle |x - 4| < \frac{1}{2} \implies -\frac{1}{2} < x - 4 \implies -28 < 56x - 224 \implies \frac{77}{4} < 56x - \frac{707}{4}$

that means

\displaystyle \begin{align*}56x - \frac{707}{4} &< 7x^2 - 63 \\ \frac{77}{4} &< 7x^2 - 63 \\ \frac{\sqrt{77}}{2} &< \sqrt{7x^2 - 63} \end{align*}

So finally

$\displaystyle \sqrt{2} + \frac{\sqrt{77}}{2} < \sqrt{2x^2 - 7x + 3} + \sqrt{7x^2 - 63}$

and that means we define

$\displaystyle M = \frac{1}{\sqrt{2} + \frac{\sqrt{77}}{2}} = \frac{2}{2\sqrt{2} + \sqrt{77}}$

13. well, see i'm lazy. i don't want to do all that horrid algebra.

we have some horrendously ugly fraction 1/(ab) that we would like to be less than some constant M.

now, it would be nice if we could restrict x so that a > 1, b > 1. but a = √(x-3), and for a > 1, a^2 = x-3 must be > 1, and so x > 4.

but we can't insist x be larger than 4, because then we are only taking a one-sided limit.

on the other hand, we really only want to consider x > 3, or else √(x-3) isn't even defined. so we need to pick some 0 < δ < 1.

any value would do, 3/4 might have worked out better, i just "split the difference".

now if we choose δ < 1/2, it is clear that 3.5 < x < 4.5, so 0.5 < x - 3 < 1.5, so

√2/√3 < 1/√(x-3) < √2. all i care about is the upper bound (we will be taking the absolute value,

and we want to bound THAT from above). so i have 1/a bounded above: 1/(ab) < M(1/b) (as long as b > 0). i have M = √2.

so now i'd like to find an upper bound for 1/b. it would be nice if 1 was an upper bound, so that 1/(ab) < M(1/b) < M (provided b > 0).

if b > 1, i know that b > 0, so my reasoning above is justified.

b, in this case, is: √(2x-1) + √(7x - 21), and since x > 3.5 > 3, the term √(7x - 21) > 0.

so b = √(2x-1) + √(7x - 21) > √(2x-1). now x > 3.5, so its a fortiori > 1. so 2x > 2 (in fact, it's bigger than 7).

so 2x-1 > 1 (in fact, it's bigger than 6). so √(2x-1) > 1 (in fact, it's bigger than √6, which is around 2.45, but i don't care, that's still bigger than 1).

so b > 1. that's all i need to know. i don't need to know HOW much bigger than 1 it is, and i don't care.

since b > 1 > 0, 0 < 1/b < 1. we don't need to find an optimal value for M (which is the minimal, NOT the maximal value). we just need to find one that works.

√2 works. now we have that 1/(ab) < √2.

but what we actually have to bound is |x-4||-5||1/(ab)| = |x-4|(5/(ab)) (since ab is positive. we just saw that b is, and a has to be, or else its undefined).

and |x-4|(5/(ab)) < |x-4|5√2. if we wish this to be < ε, for any ε > 0, we should choose δ = min(1/2, ε/(5√2)).

now, why is this preferable to the arduous algebra above? well, it's easier to get to. sure, my "M" is bigger,

but that means my "δ" is smaller, so i might be way less than ε than i need to be, but that dosn't matter.

i just need to find SOME δ > 0 that works (there exists...) to get the absolute value of the difference of my expression and the limit under ε.

yes, it's a crude bound, but these aren't equations, they're inequalities.

14. Alright, I gave in and looked at my solutions manual.

They're going from:

$|\frac{-5(x-4)}{ (\sqrt{x-3} )(\sqrt{2x-1}-\sqrt{7x-21} )} |$

To

$|x-4| \times \frac{5} }{ (\sqrt{x-3)(\sqrt{2x-1}-\sqrt{7x-21} )}$

Why is it legal for me to just take the ABV out of that whole division operation? The abv would have previously been of the denominator too, but now it isn't, shouldn't that matter?

15. No, $\displaystyle \frac{ab}{c} = a\left(\frac{b}{c}\right)$

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