Proofs for limits, feels unfamilar

[1MileCrash]

Firstly, I find the math syntax on this board incredibly difficult to use, so bear with me. Using any symbol makes the text appear on the next line... I don't know if it is my browser, or what, but I tried to make due. Sorry.

Homework Statement

Construct an "epsilon minus delta" proof for each limit fact.

limit of x approaching 5 of:

(x^2 - 25) / (x - 5) = 10

Homework Equations

-

The Attempt at a Solution

Pre-work:

( x^2 - 25 ) / ( x - 5 )
=
( x + 5 )


0 < |x - 5| < (delta) => | ( x + 5 ) - 10 | < (epsilon)
| ( x - 5 ) | < (epsilon)

Proof:

| ( x^2 -25 ) / ( x - 5 ) | = | x - 5 | < (delta) = (epsilon)





Obviously a very simple problem since the limit statement and function are essentially the same, but I don't remember the exact "content" people are looking for when they want limits proven.

 

[Dick]

That's exactly what you want. If f(x)=(x^2-25)/(x-5) then if you pick any epsilon and set delta=epsilon then |x-5|<delta implies that |f(x)-10|<epsilon.

 

[1MileCrash]

Thank you sir, are most professors going to dislike me simplifying the function outside of the "preliminary analysis" most of them call for?

For example:

limit x -> 1 of
( 10x^3 - 26x^2 + 22x - 6 ) / ( x - 1 )^2

f(x) = 10x - 6

Preliminary:

0 < | x - 1 | < delta => | (10x - 6) - 4 | < epsilon
=> | (10x - 10 | < epsilon
=> 10 | x - 1 | < epsilon
=> | x - 1 | < epsilon / 10

Proof:

| ( 10x - 6 ) - 4 | = | 10x - 10 | = 10 | x - 1 | < 10(delta) = epsilon

Can I just treat the problem as if they gave me 10x - 6 from the get-go, or should I show my work for that in the preliminary work? I'm always careful about these things because it's only partially knowing the math.

 

[Dick]

(10x^3 - 26x^2 + 22x - 6)/(x-1)^2 is the SAME thing as 10x - 6 UNLESS x=1. And if you are evaluating the limit as x->1 you don't care about the value of the function at x=1. Not even a little. If you can convince your professor by stating you understand that the functions aren't the SAME, but they are for the purpose of evaluating the limit, then I really don't think any reasonable person could possibly object.

 

[Slats18]

IMO, you should always show your working, so you should add how you simplified your numerator, as it's not entirely obvious by inspection =)

 

[Dick]

IMO, you should always show your working, so you should add how you simplified your numerator, as it's not entirely obvious by inspection =)

Sure, sure. You should probably also show you factored (10x^3 - 26x^2 + 22x - 6) to (x-1)^2*(10x-6). Wouldn't hurt for the purposes of the presentation.

 

[1MileCrash]

Alright guys, I worked on this one for a good 15 minutes today... I don't have my attempt at a solution, it's huge and in my notes, but I got stuck.

lim
x -> 4 f(x) = sqrt(7)

f(x) = [sqrt(2x-1)] / [sqrt(x-3)]

Any takers on this doozie?

 

[HallsofIvy]

What is the question? You seem to have written this reversed: if f(x)= sqrt(2x-1)/sqrt(x-3) then limit as x-> 4 of f(x) is sqrt(7).

But what are you asking? If you are just asking how to show that, since sqrt is a continuous function, this is simply sqrt(2(4)- 1))/sqrt(4- 3)= sqrt(7)/sqrt(1)= sqrt(7).

If you are asking how to prove it from the definition, then you are going to have to look at
$$\frac{\sqrt{2x-1}}{\sqrt{x-3}}- \sqrt{7}= \frac{\sqrt{2x-1}- \sqrt{7x- 21}}{\sqrt{x- 3}}$$
and I would recommend multiplying both numerator and denominator by the complement,
$$\sqrt{2x-1}+ \sqrt{7x- 21}$$