Limit 2

doodlenoodle · **Joined:** Sun, 12 Feb 2012 02:16:54 UTC **Posts:** 64

Evaluate the given limits and prove your conclusion.

$$\lim_{x\to1}\frac{x^2-x-2}{2x-3}$ = $|x-1||\frac{x-4}{2x-3}|$

Trying to get the upper bound for the second part, my professor told me to ch0ose |x-1|<1/4 instead of my original |x-1|<1, from here I know how to proceed but I dont understand why he had this restriction. In general when can you not put a bound say 1 on |x-s|<1

Shadow · **Posted:** Wed, 21 Mar 2012 04:09:43 UTC

Post new questions in new topics. Topic split.

doodlenoodle · **Joined:** Sun, 12 Feb 2012 02:16:54 UTC **Posts:** 64

Yes sir. And yeah, I think I get my question now. It's because x cannot equal 3/2, and saying |x-1|<1 implies that x could possibly equal 3/2, so by saying |x-1|<1/4, the maximum value x could be is 1.25, which is less than 1.5, so it's safe to say.

outermeasure · **Posted:** Wed, 21 Mar 2012 04:15:04 UTC

doodlenoodle wrote:

Evaluate the given limits and prove your conclusion.

$$\lim_{x\to1}\frac{x^2-x-2}{2x-3}$ = $|x-1||\frac{x-4}{2x-3}|$

Trying to get the upper bound for the second part, my professor told me to ch0ose |x-1|<1/4 instead of my original |x-1|<1, from here I know how to proceed but I dont understand why he had this restriction. In general when can you not put a bound say 1 on |x-s|<1

What is this nonsense? The LHS $\displaystyle\lim_{x\to1}\frac{x^2-x-2}{2x-3}$ is a number, the RHS $\displaystyle\lvert x-1\rvert\left\lvert\frac{x-4}{2x-3}\right\rvert$ is a nonconstant function of x, and you are asserting they are equal?

doodlenoodle · **Joined:** Sun, 12 Feb 2012 02:16:54 UTC **Posts:** 64

No no no, must have worded my question poorly. I was meaning to ask how to find the bounds on the

$|\frac{x-4}{2x-3}$ term, and now I see why I must chose |x-1|<1/4, and thus |x-4| is bounded by -13/4<x-4<-11/4 and that

|2x-3| is bounded by -5/2<2x-3<-1/2, I would take the fraction then to be $|\frac{x-4}{2x-3}|$ <13/2

Shadow · **Posted:** Wed, 21 Mar 2012 04:26:10 UTC

doodlenoodle wrote:

No no no, must have worded my question poorly. I was meaning to ask how to find the bounds on the

$|\frac{x-4}{2x-3}$ term, and now I see why I must chose |x-1|<1/4, and thus |x-4| is bounded by -13/4<x-4<-11/4 and that

|2x-3| is bounded by -5/2<2x-3<-3/2, I would take the fraction then to be $|\frac{x-4}{2x-3}|$ <13/6

I don't even see how you are relating the bit of functions on each side, I still object to what thinking either side has to do with one another.

doodlenoodle · **Joined:** Sun, 12 Feb 2012 02:16:54 UTC **Posts:** 64

For a Delta Epsilon proof, I would have Let e>0 be given, we need to find d>0 such that if 0<|x-1|<d, then |f(x)-2|<e.

Do what I did,

Choose d=min{1/4,(13/2)e}. Then if |x-1|<d, then |f(x)-2|<e.

outermeasure · **Posted:** Wed, 21 Mar 2012 04:39:51 UTC

doodlenoodle wrote:

For a Delta Epsilon proof, I would have Let e>0 be given, we need to find d>0 such that if 0<|x-1|<d, then |f(x)-2|<e.

Do what I did,

Choose d=min{1/4,(13/2)e}. Then if |x-1|<d, then |f(x)-2|<e.

Well, a proof is, at a start, a coherent prose. Maybe you need to improve in that department first instead of making us second guessing what you might mean to say as opposed to what you said. You haven't shown what you did between $\displaystyle\lim_{x\to 1}\frac{x^2-x-2}{2x-3}$ and $\displaystyle\lvert x-1\rvert\left\lvert\frac{x-4}{2x-3}\right\rvert$ except an equal sign between them, and so "(D)o what I did" is a definite no-no.

doodlenoodle · **Joined:** Sun, 12 Feb 2012 02:16:54 UTC **Posts:** 64

Definitely over-complicated things. My original question was why couldn't I bound |x-1|<1. I did not mean to answer the question or act as if this was a proof I'd turn in. Also the original post was in my previous Delta-Epsilon topic so my preface was thrown off. My mistake though, I will try and be more clear.

Shadow · **Posted:** Wed, 21 Mar 2012 05:07:02 UTC

doodlenoodle wrote:

Definitely over-complicated things. My original question was why couldn't I bound |x-1|<1. I did not mean to answer the question or act as if this was a proof I'd turn in. Also the original post was in my previous Delta-Epsilon topic so my preface was thrown off. My mistake though, I will try and be more clear.

Yes, that's fine, but we still have no idea why you would want to talk about |x-1| in the first place, much less why you would want it less than 1 or 1/4 or something else entirely. Even knowing this is epsilonics, doesn't explain what you're talking about for this particular problem, which you always need to explain, otherwise we have little possibility of understanding you.

outermeasure · **Posted:** Wed, 21 Mar 2012 05:10:52 UTC

doodlenoodle wrote:

Definitely over-complicated things. My original question was why couldn't I bound |x-1|<1. I did not mean to answer the question or act as if this was a proof I'd turn in. Also the original post was in my previous Delta-Epsilon topic so my preface was thrown off. My mistake though, I will try and be more clear.

Huh? This question has nothing to do with your previous thread ( $\displaystyle\lim_{x\to -1}\frac{1}{(x^2+1)^{1/2}}=\frac{1}{2^{1/2}}$ has nothing to do with $\displaystyle\lim_{x\to 1}\frac{x^2-x-2}{2x-3}$ or $\displaystyle\lvert x-1\rvert\left\lvert\frac{x-4}{2x-3}\right\rvert$ ), which is the reason why it was split off.

S.O.S. Mathematics CyberBoard

Limit 2

Who is online