Help with fundamental concept of calculus; limits.

  • #1

Main Question or Discussion Point

I'm a high school student and am a junior right now. I like physics, and quantum theory, perhaps because it will be our future.


Therefore, in order to understand it, I need to know calculus. Since I don't have homework, I don't work out the problems. I try to understand the critical concepts, then perhaps try and solve it.

Here's a problem I'm having with basic limits (Still can't figure out where to get the limit sign and stuff)

Where a = x and a is some given x value.

Then lim x -> a = f(x)

Or something like that. The point is this; I looked at the equation and could not understand one sided limits.

Here is something I don't understand. There are two asypthopes, horiztonal and vertical. As the function approached 3, I saw according to looking on the y axis..it went PAST 3..but looking on the x axis..it still had no reached 3.

So I think I'm looking at it wrong. Does a have to always equal a x value? Or can it equal a = f(x,y) later? (Mutivarible calc?)
 

Answers and Replies

  • #2
arildno
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This is why you SHOULD do homework. Your ideas are so muddled here, that it is impossible to figure out what you are actually asking about.
 
  • #3
It was off the top of my head.Oh, and the book is "Calculus for Dummies" they don't give you problems.:(

Ok:

Consider the limit of the function f(x) = (x+2)(x - 5)/(x - 3)(x+1)

(Could you guys graph this?)

It says:

"As x approachs 3 from the left, (I see the arrow going towards the hortizontal asymptote on Quadrant 1 on the xy axis. which makes sense, as it never gets to 3.I understand this..), f(x) goes up to infinity; (I assume infintiy is on the same parabola of the curve with two arrows..as one side approachs 3 the other side approachs infinity? I think I understand this too..) and as x approachs 3 from the right, f(x) goes down to negative infinity. (Now, I DO NOT understand this. The problem lies in that they list 3 paraboles. With two vertical asymptotes || to each other. The left vertical asymptote intersects the x axis at (-1,1) while the other vertical asymptote intersects the x axis on Quadrant 1 at (3,0). There is also a hortizontal asymptote apparently going from negative infinity to postitive infinity precisely at (0,1). There is one more parabola that is between all three parabolas. The midpoint of the curve appears to be at approx. 3 while the two arrows seem to go up to postive infinity).
 
  • #4
JasonRox
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Calculus for Dummies is for dummies that are dumb enough to buy one. You got ripped.

Buy a real textbook. James E. Stewart is a good author. Look for him.

The best way that I can help you is to actually write the question.

When they say from the right, they are talking about 3.1, 3.01, 3.001 and so on. From the left is 2.9 and so on. It can be negative infinity on one side and positive on the other. Look at 1/x.

Calculus for Dummies is a book to accompany those who don't get it. They don't actually teach you anything. They just repeat the same damn thing in the textbook, in hopes that after reading it twice you get it.

I'm not trying to discourage you at all. Get a textbook. No fancy smancy entertainment book.
 
  • #5
graphic7
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If you'd like a plot:

http://graphical.shacknet.nu/temp/plot.jpg [Broken]

Just follow the function with your finger, starting with the negative direction. As your "approaching" 3 from the left where are you going? Remember, approaching doesn't always mean 'touching.'

It's all about where your finger is headed. Make jumps until you get to the point where your approaching and then start observing where your finger is going. If it goes off the image, that should give you a hint. :biggrin:

And one more very important thing about limit, althought it doesn't apply in this question. Does the limit exist as x approaches 3? In order to evaluate this question you must take the limit from both the left and right. If they are equal then, in fact, the limit does exist. However, let's suppose the limits from the left and right directions are different, then, the limit does not exist. In this problem, try figuring out the limit as x approaches 3.
 
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  • #6
shmoe
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QuantumTheory said:
(Now, I DO NOT understand this. The problem lies in that they list 3 paraboles.
The pieces of the graph aren't "parabolas". Parabolas don't have asymptotes.

QuantumTheory said:
With two vertical asymptotes || to each other. The left vertical asymptote intersects the x axis at (-1,1)
It can't intersect the x-axis at (-1,1), since this isn't a point on the x-axis.

To find

[tex]\lim_{x\rightarrow 3^+}\frac{(x+2)(x - 5)}{(x - 3)(x+1)}[/tex]

Since

[tex]\frac{(x+2)(x - 5)}{(x+1)}=(5)(-2)/(4)=-5/2[/tex] and [tex](3-3)=0[/tex]

Or function looks kinda like [tex]\frac{-5/2}{0}[/tex]

Of course in the limit we never actually divide by zero, since x never actually reaches 3. However, we are dividing something that's very close to -5/2 by something that get's closer and closer to 0 so the end result is something that get's arbitrarily large in absolute value and the limit will be plus or minus infinity.

To determine which way it's going, consider the sign of (x-3) in the limit. You're appraoching 3 from the right, which means x is always larger than 3. Thus, x-3 is always positive.

So as x approaches 3 from the right we have something close to -5/2 divided by something positive that's approaching 0. We conclude that

[tex]\lim_{x\rightarrow 3^+}\frac{(x+2)(x - 5)}{(x - 3)(x+1)}=-\infty[/tex]

You really need to work out problems rather than just try to understand the concepts. If your Calculus for Dummies book doesn't have problems then get another book that does.
 
  • #7
shmoe said:
The pieces of the graph aren't "parabolas". Parabolas don't have asymptotes.



It can't intersect the x-axis at (-1,1), since this isn't a point on the x-axis.

To find

[tex]\lim_{x\rightarrow 3^+}\frac{(x+2)(x - 5)}{(x - 3)(x+1)}[/tex]

Since

[tex]\frac{(x+2)(x - 5)}{(x+1)}=(5)(-2)/(4)=-5/2[/tex] and [tex](3-3)=0[/tex]

Or function looks kinda like [tex]\frac{-5/2}{0}[/tex]

Of course in the limit we never actually divide by zero, since x never actually reaches 3. However, we are dividing something that's very close to -5/2 by something that get's closer and closer to 0 so the end result is something that get's arbitrarily large in absolute value and the limit will be plus or minus infinity.


So as x approaches 3 from the right we have something close to -5/2 divided by something positive that's approaching 0. We conclude that

[tex]\lim_{x\rightarrow 3^+}\frac{(x+2)(x - 5)}{(x - 3)(x+1)}=-\infty[/tex]

You really need to work out problems rather than just try to understand the concepts. If your Calculus for Dummies book doesn't have problems then get another book that does.

I don't understand what you're saying. It confuses me. Please don't tell me to "get a calculus book", it is very confusing for me.

I will try to work out problems. But I don't understand this. The book said the limit of both left and right limits do not exist because they are not equal. But he said both left and right limits can equal a finite or infinite value?

[q]To determine which way it's going, consider the sign of (x-3) in the limit. You're appraoching 3 from the right, which means x is always larger than 3. Thus, x-3 is always positive.[/q]

How does that tell me it's approaching 3 from the right?

The author said each piecewise function can have a infinite or finite value, but them together must be equal in order for the limit to exist. Why is this??

:confused:
 
  • #8
If it is lim x -> 3+ then according to the graph to what a poster posted doesn't x -> 3+ equal infinity and f(3) equals negative infinity??

Does x->3+ mean to look at the graph I start at "3" (By "3" I mean very close to 3, since you can't get to 3) and go up on the curve??

But doesn't that equal infinity if its going up?
 
  • #9
shmoe
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QuantumTheory said:
How does that tell me it's approaching 3 from the right?
It was a right hand limit. Your function has discontinuities at x=3 and x=-1. In order to find out what the graph looks like near these discontinuities you have to evaluate the one sided limits at these points. I was working through the right handed limit at 3 for you. Loosely speaking, [tex]{\mbox\lim_{x\rightarrow 3^{+}}f(x)[/tex] is what your funcion is approaching as you evaluate it at points larger than 3 that get closer and closer to 3.

QuantumTheory said:
The author said each piecewise function can have a infinite or finite value, but them together must be equal in order for the limit to exist. Why is this??
The (2-sided) limit of a function at a point exists if and only if both of the one-sided limits exist and are equal. In symbols:

[tex]\lim_{x\rightarrow a}f(x)=L[/tex]

if and only if

[tex]\lim_{x\rightarrow a^{-}}f(x)=L[/tex] and [tex]\lim_{x\rightarrow a^{+}}f(x)=L[/tex]

Now, if you look closely at the definition of the limit you'll see it takes into account values of x on either side of a, while the one-sided limits only reflect the values of x on one side. Your book should go into some detail on this, if it doesn't, get another book (I mean this in all seriousness and for your own benefit, if you book is dodgy your understanding will be dodgy).

QuantumTheory said:
If it is lim x -> 3+ then according to the graph to what a poster posted doesn't x -> 3+ equal infinity and f(3) equals negative infinity??
f(3) is not defined, it doesn't equal anything.

QuantumTheory said:
Does x->3+ mean to look at the graph I start at "3" (By "3" I mean very close to 3, since you can't get to 3) and go up on the curve??
Well, your starting point is correct, but you follow along the curve as you head to the left. So the curve is going down as you head left.

QuantumTheory said:
But doesn't that equal infinity if its going up?
It's going down. You're moving left. Think about this, if you are standing on the side of a hill and facing "uphill" if you turn around you are now facing "downhill". I can't be sure, but I think you are walking backwards while facing "uphill" and declaring you are going "uphill".
 
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  • #10
shmoe said:
It was a right hand limit. Your function has discontinuities at x=3 and x=-1. In order to find out what the graph looks like near these discontinuities you have to evaluate the one sided limits at these points. I was working through the right handed limit at 3 for you. Loosely speaking, [tex]{\mbox\lim_{x\rightarrow 3^{+}}f(x)[/tex] is what your funcion is approaching as you evaluate it at points larger than 3 that get closer and closer to 3.



The (2-sided) limit of a function at a point exists if and only if both of the one-sided limits exist and are equal. In symbols:

[tex]\lim_{x\rightarrow a}f(x)=L[/tex]

if and only if

[tex]\lim_{x\rightarrow a^{-}}f(x)=L[/tex] and [tex]\lim_{x\rightarrow a^{+}}f(x)=L[/tex]

Now, if you look closely at the definition of the limit you'll see it takes into account values of x on either side of a, while the one-sided limits only reflect the values of x on one side. Your book should go into some detail on this, if it doesn't, get another book (I mean this in all seriousness and for your own benefit, if you book is dodgy your understanding will be dodgy).



f(3) is not defined, it doesn't equal anything.



Well, your starting point is correct, but you follow along the curve as you head to the left. So the curve is going down as you head left.



It's going down. You're moving left. Think about this, if you are standing on the side of a hill and facing "uphill" if you turn around you are now facing "downhill". I can't be sure, but I think you are walking backwards while facing "uphill" and declaring you are going "uphill".

I see what you mean. Because it never equals 3. You could make a series as follows:

3.01, 3.001, 3.0001, 3.00001, 3.000001, ....., 3.000000000000000000000001


And since 3 does exist on the axis. We are talking as [tex]\lim_{x\rightarrow 3}[/tex], it never becomes 3 but it becomes infinitesimally close to 3.
 

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