20180919, 01:28  #34  
Jul 2003
Behind BB
3341_{8} Posts 
Quote:
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. 

20180919, 08:26  #35 
Banned
"Luigi"
Aug 2002
Team Italia
2·2,417 Posts 
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. 
20180920, 08:11  #36 
Romulan Interpreter
"name field"
Jun 2011
Thailand
2^{4}·613 Posts 
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. Last fiddled with by LaurV on 20180920 at 08:18 Reason: spacing (grrr) 
20180920, 12:11  #37  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2^{4}·13·53 Posts 
Quote:
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 ... 

20180920, 23:32  #38  
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{2}·3·37 Posts 
Quote:
Quote:
I'm still surprised that we (the class) are learning about limits without learning about the epsilondelta 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 epsilondelta 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: 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? 

20180921, 06:56  #39  
Romulan Interpreter
"name field"
Jun 2011
Thailand
2^{4}×613 Posts 
Quote:


20180921, 08:28  #40  
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
2^{4}·13·53 Posts 
Quote:
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. 

20180921, 23:57  #41  
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
110111100_{2} Posts 
Quote:
Quote:
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...\[2(1\sin ^2 x) = 13 \sin x + 5\]\[22\sin^2x = 13 \sin x + 5\]\[2\sin^2x + 13 \sin x + 3 = 0\]So \(\sin x = a\)...\[2a^2+13a+3\]\[a = \dfrac{13 \pm \sqrt{13^24(2)(3)}}{2(2)}\]\[a = \dfrac{13 \pm \sqrt{16932}}{4}\]\[a = \dfrac{13 \pm \sqrt{145}}{4}\]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...? 

20180922, 03:14  #42 
Jun 2003
3·11·157 Posts 

20180924, 21:50  #43 
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{2}·3·37 Posts 
Oops...
We covered some more on limits today; last week we went over a problem involving a graph and a vertical asymptote, like this: 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 \(\lim_{x\rightarrow 0}(x^24) = \lim_{x\rightarrow 0}x^2  \lim_{x\rightarrow 0}4 = (\lim_{x\rightarrow 0}x)^2  \lim_{x\rightarrow 0}4\). 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 xvalue 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? 
20180925, 01:08  #44 
"Curtis"
Feb 2005
Riverside, CA
37×137 Posts 
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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