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Old 2018-09-19, 01:28   #34
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Originally Posted by jvang View Post
The teacher said that the infinite limits correspond to vertical asymptotes, but would the other forms of limits technically be vertical asymptotes, too? Maybe my definition of asymptotes is misguided

We also covered graphs that oscillate between positive and negative while approaching the limit value, which means that we couldn't find the limit. An example was \(f(x) = \sin \dfrac{\pi}{x}\). Is there a way to find the limit of this function?

Guess this is going to take a while...
Limits are a major building block for the upcoming concepts in a calculus course. If I were to write a calculus textbook, it would have five chapters. One of those chapters would be titled: limits.

Infinite limits don't have to be vertical asymptotes. Consider f(x) = x^3 and the limit as x approaches infinity or g(x) = -x^3.

For the sin(1/x) example, there is no limit. Careful study of the definition of a limit should make this clear.
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Old 2018-09-19, 08:26   #35
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One thing I learnt about limits at the college was "always use the definition of limit to show if a function is limited or not". In other words, either picture the graph in your mind, or use enough large M or enough small epsilons to find a value over/under which the definition is definitely true.
In 99% of the cases, if such value doesn't exist or can't b reached definitely, the limit does not exist.
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Old 2018-09-20, 08:11   #36
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Sure, you need some kind of "rigorous" proof, neighborhoods, whatever, otherwise is tricky, "by eyes" you can run in all kind of strange results, I still remember one of our teacher favorite demonstrations of the fact that 1==2. It went like that: take an equilateral triangle with size s (only the laterals, no interior) and bend it in half, then continue to bend it in half on the dotted lines, a million times, a billion times, an infinite number times, like in this photo:

Click image for larger version

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Then it is clear the blue broken line will, at the limit, be equal with the red line, no question about it, its thickness is getting to zero, and its length stays bounded by s, the length of the red lateral. Right? Are you with me still?

But now, if you calculate the length of the blue broken line, in the first image it is s+s=2s, in the second it is s/2+s/2+s/2+s/2= 4 times s/2, so it is 2s, in the third it is 8 times s/4, equal 2s, and so on, so the length of the blue line is always 2s. It is never getting smaller. Therefore, at the limit, you have s=2s, and then you simplify to get 1==2, hehe.

Last fiddled with by LaurV on 2018-09-20 at 08:18 Reason: spacing (grrr)
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Old 2018-09-20, 12:11   #37
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Quote:
Originally Posted by LaurV View Post
Sure, you need some kind of "rigorous" proof, neighborhoods, whatever, otherwise is tricky, "by eyes" you can run in all kind of strange results, I still remember one of our teacher favorite demonstrations of the fact that 1==2. It went like that: take an equilateral triangle with size s (only the laterals, no interior) and bend it in half, then continue to bend it in half on the dotted lines, a million times, a billion times, an infinite number times, like in this photo:

Attachment 19071

Then it is clear the blue broken line will, at the limit, be equal with the red line, no question about it, its thickness is getting to zero, and its length stays bounded by s, the length of the red lateral. Right? Are you with me still?

But now, if you calculate the length of the blue broken line, in the first image it is s+s=2s, in the second it is s/2+s/2+s/2+s/2= 4 times s/2, so it is 2s, in the third it is 8 times s/4, equal 2s, and so on, so the length of the blue line is always 2s. It is never getting smaller. Therefore, at the limit, you have s=2s, and then you simplify to get 1==2, hehe.
A similar proof is that sqrt(2) = 2.

Draw right angled triangle with sides 1, 1 and sqrt(2) --- the latter follows by Pythagorus. Now approximate the hypotenuse by a series of lines parallel to the other two sides. The length of this path is 2, independent of the number of lines in the series. Let the number of lines tend to infinity ...
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Old 2018-09-20, 23:32   #38
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Quote:
Originally Posted by LaurV View Post
Sure, you need some kind of "rigorous" proof, neighborhoods, whatever, otherwise is tricky, "by eyes" you can run in all kind of strange results, I still remember one of our teacher favorite demonstrations of the fact that 1==2. It went like that: take an equilateral triangle with size s (only the laterals, no interior) and bend it in half, then continue to bend it in half on the dotted lines, a million times, a billion times, an infinite number times, like in this photo:

Then it is clear the blue broken line will, at the limit, be equal with the red line, no question about it, its thickness is getting to zero, and its length stays bounded by s, the length of the red lateral. Right? Are you with me still?

But now, if you calculate the length of the blue broken line, in the first image it is s+s=2s, in the second it is s/2+s/2+s/2+s/2= 4 times s/2, so it is 2s, in the third it is 8 times s/4, equal 2s, and so on, so the length of the blue line is always 2s. It is never getting smaller. Therefore, at the limit, you have s=2s, and then you simplify to get 1==2, hehe.
It seems weird; but the length of each individual piece as you divide by two tends towards 0. Which means we are dividing by 0?

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Originally Posted by xilman View Post
A similar proof is that sqrt(2) = 2.

Draw right angled triangle with sides 1, 1 and sqrt(2) --- the latter follows by Pythagorus. Now approximate the hypotenuse by a series of lines parallel to the other two sides. The length of this path is 2, independent of the number of lines in the series. Let the number of lines tend to infinity ...
What do you mean by a "series of lines"? I'm having trouble visually representing what you're talking about here...

I'm still surprised that we (the class) are learning about limits without learning about the epsilon-delta definition. I'm pretty sure that we will cover it, but it's annoying listening to the teacher's ELI5 definition: "A limit is a 'wannabe' value on the function's graph, which it 'wants' to get to but can't."

However, I really dislike anthropomorphizing (this is the word, but I want to spell it "anthropomorphisizing"?) mathematical concepts and other things that we learn about. We (at least the school district around here) do it a lot in science and math classes

The Khan Academy video gave a pretty decent definition of the epsilon-delta limit concept. IIRC the idea is that, for a given epsilon, you add and subtract the epsilon value and the \(y\)-value of the proposed value of the limit. The job is to find a delta value to add/subtract with the \(x\)-value of the limit, where \(\delta \leq \epsilon\). This is better described in the below picture:

Click image for larger version

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I forget anything else regarding this, since I'm not good at learning from videos. I'm 90% sure I'm missing something regarding how we prove that this is true for any epsilon; I think it seems logically plausible, but not necessarily mathematically rigorous. What important properties did I miss?
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Old 2018-09-21, 06:56   #39
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Quote:
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It seems weird; but the length of each individual piece as you divide by two tends towards 0. Which means we are dividing by 0?
Not really, s is a constant... The apparent paradox comes from the fact that in reality, you have two functions that "fit in the same 2D container", which (container) has length s and width 0, and both functions are continue in any point, but the first function is also continuous derivable in any point and has the length s, contrary to the second which is not derivable in any point, and it has the length 2s (or any number, by the way, as Xilman said). The existence of a function which is always continue but never derivable (very contrary to our 'normal' intuition) was proved long ago by Weierstrass, of whose name it got.
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Old 2018-09-21, 08:28   #40
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What do you mean by a "series of lines"? I'm having trouble visually representing what you're talking about here...
Draw the triangle with the hypotenuse slanting, one side vertical and the other horizontal.

Approximate the hypotenuse by a staircase of lines alternately horizontal and vertical. Let the number of steps in the staircase tend to infinity. The length of the staircase, that is the sum of the vertical risers and horizontal steps is always 2 units, independent of the number of steps.

I repeat my earlier advice: if you can't visualize something try making a drawing with pencil and paper.
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Old 2018-09-21, 23:57   #41
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Originally Posted by LaurV View Post
Not really, s is a constant... The apparent paradox comes from the fact that in reality, you have two functions that "fit in the same 2D container", which (container) has length s and width 0, and both functions are continue in any point, but the first function is also continuous derivable in any point and has the length s, contrary to the second which is not derivable in any point, and it has the length 2s (or any number, by the way, as Xilman said). The existence of a function which is always continue but never derivable (very contrary to our 'normal' intuition) was proved long ago by Weierstrass, of whose name it got.
That seems interesting, but I have no concept of what a derivative is, so this is like reading a different language

Quote:
Originally Posted by xilman View Post
Draw the triangle with the hypotenuse slanting, one side vertical and the other horizontal.

Approximate the hypotenuse by a staircase of lines alternately horizontal and vertical. Let the number of steps in the staircase tend to infinity. The length of the staircase, that is the sum of the vertical risers and horizontal steps is always 2 units, independent of the number of steps.

I repeat my earlier advice: if you can't visualize something try making a drawing with pencil and paper.
Ohhh, that makes more sense. I was really confused... I was trying to draw something completely different!

The trigonometry problem earlier in the thread did indeed have a typo. This:\[2 \cos x = 13 \sin x + 5\] was supposed to be \[2\cos^2 x = 13 \sin x + 5\]Let's see how it simplifies...\[2(1-\sin ^2 x) = 13 \sin x + 5\]\[2-2\sin^2x = 13 \sin x + 5\]\[2\sin^2x + 13 \sin x + 3 = 0\]So \(\sin x = a\)...\[2a^2+13a+3\]\[a = \dfrac{-13 \pm \sqrt{13^2-4(2)(3)}}{2(2)}\]\[a = \dfrac{-13 \pm \sqrt{169-32}}{4}\]\[a = \dfrac{-13 \pm \sqrt{145}}{4}\]I'm pretty sure that this worked out to be a neat answer in class, but thinking back on the review we ended up with \(2\sin^2x + 13 \sin x + 7 = 0\), which factors into \((2\sin x + 7)(\sin x + 1)\). Then you just solve for each \(\sin x\). But that seems wrong and I don't think that I made any mistakes in the above equations...?
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Old 2018-09-22, 03:14   #42
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\(2\sin^2x + 13 \sin x + 7 = 0\), which factors into \((2\sin x + 7)(\sin x + 1)\).
(2x+7)*(x+1) = 2x^2+9x+7
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Old 2018-09-24, 21:50   #43
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(2x+7)*(x+1) = 2x^2+9x+7
Oops...

We covered some more on limits today; last week we went over a problem involving a graph and a vertical asymptote, like this:

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Where both sides of the graph tend towards positive infinity as x approaches -1. Our teacher said that the limit does not exist here, but I asked if it would be incorrect to answer the problem with \(\lim_{x\rightarrow -1}f(x) = \infty\). Is this correct, or does the limit not exist?

We also covered some rules regarding limits, which basically seem like treating \(\lim\) as a function. Things such as \(\lim_{x\rightarrow 0}(x^2-4) = \lim_{x\rightarrow 0}x^2 - \lim_{x\rightarrow 0}4 = (\lim_{x\rightarrow 0}x)^2 - \lim_{x\rightarrow 0}4\). We're using the fact that \(\lim_{x\rightarrow c}x = c\) and that \(\lim_{x\rightarrow c}\) of any constant returns that constant. The "special" rules regarding these limits lets us break up functions into these base forms, then evaluate from there. However, it seems like we could just plug in whatever x-value we're looking for would work in most cases?

Man, my limit notation is still messed up! How was it working earlier in the thread?
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Old 2018-09-25, 01:08   #44
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For the graph you've exhibited, the limit as x approaches -1 is +infinity.
In order to say a limit is +inifinity, we must be able to find a c for any given M such that f(x) > M for all x between c and -1.
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