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Geek Culture / I've been trying to learn calculus and I'm having a little trouble

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easter bunny
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Posted: 25th Aug 2014 03:51 Edited at: 25th Aug 2014 03:56
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So I've decided to give calculus a shot and I'm going through this course: http://www.coursera.org/learn/calculus1/

I'm up to a quiz and here are the limits I have to figure out:

sin(2x)/8x
sin(9x)/6x
sin(-2x)/x

each as x approaches 0


But I'm not exactly sure how I'm supposed to figure them out, am I just supposed to 'brute force' them? ie. punch in a few numbers approaching it from each side and interpolate from that to what the answer is?
The easy way which I found is the type the equation into Google, it will graph the whole thing for you and it's quite easy to see what the answer will be (this is not really cheating, it's just graphing the equation the easy way )
ie. http://www.google.com.au/search?q=sin%28-2x%29%2Fx

Obviously the f(x) as x approaches 0 is -2

The problem is that I don't think I'm doing this the way I'm supposed to


Can anybody help me out?

I think I'm supposed to be using the limit rules (limit of a product is the product of the limits providing the limits exists, limit of a quotient is the quotient of the limits providing the denominator != 0, etc), but I don't see how I'm supposed to apply the laws to these problems

EDIT:
I put in the answers I got from the graph and each of them was correct, sort of.
For the third problem, it said "Well Done! Your answer, -2, is equivalent to the instructors answer of -2/1"
Each of the other answers by the instructor are expressed as fractions as well, does this have any significance?


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Libervurto
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Posted: 25th Aug 2014 14:48
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I am studying calculus too! Unfortunately I can't help with your question yet but I'm about to start a session so maybe I will in a couple of hours.

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Hockeykid
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Posted: 25th Aug 2014 16:48
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Quote: "but I don't see how I'm supposed to apply the laws to these problems "


Let me see if I can explain this.

To find the limit, you need to substitute the number that x is approaching into your equation. If the answer turns out to be undefined you use the laws of limits to break up the equation into smaller pieces and then try substituting again. (For example, I believe sin(2x)/8x can become sin(2x) * 1/(8x) however it needs to be broken down even more).

Alternatively, if you know how to take the derivative you can you L'Hospital's rule (this works in most cases, except a few). Essentially you take the derivative of the numerator, and then the derivative of the denominator and substitute again. If the answer is still undefined, you keep taking the derivative and substituting until you get an answer.

Sean

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easter bunny
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Posted: 26th Aug 2014 01:01
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So that would be something like this...
sin(2x)/8x
=
sin(2x)/2*4*x
=
lim(sin(2x)) / lim(2) * lim(4) * lim(x)
=
lim(sin(2x))/lim(2) / lim(sin(2x))/lim(4) / lim(sin(2x))/lim(4)
[limit of a product is the product of the limits providing the limits exist]

Is this something like what you're saying?


Check out this graphed equation http://www.google.com.au/search?q=%28sin%282x%29%2F2%29%2F%28sin%282x%29%2F4%29%2F%28sin%282x%29%2Fx


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Hockeykid
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Posted: 26th Aug 2014 01:47 Edited at: 26th Aug 2014 01:50
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Actually, for these problems you should be able to use this:

Lim x-> 0 ( Sin( x ) / x ) = 1

So,

Lim Sin( 2x ) / 8x =

We need to make Sin( 2x ) / 8x = 1 so we multiply by 2/2
Lim ( Sin( 2x ) / 8x ) * 2/2 =

Switch around the 2 and the 8 in the denominator to make it easier to read.
Lim ( Sin( 2x ) / 2x ) * 2/8 =

( Sin( 2x ) / 2x ) will now become 1. Here's what we are left with
Lim ( 1 ) * 2/8 =

Which just ends up being 1/4
1/4

While the Laws of Limits weren't really used in this problem, it's a good idea to memorize them because they are used in most other Limit problems.

Edit: If you did it with L'Hospital's rule it would look like this:

Lim Sin( 2x ) / 8x

Derivative of Sin( 2x ) = 2Cos( 2x )
Derivative of 8x = 8 =

2Cos( 2x ) / 8 =

2 / 8 =

1 / 4

Sean

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Libervurto
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Posted: 26th Aug 2014 10:02 Edited at: 26th Aug 2014 10:03
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Practicing my LaTex on HockeyKid's reply



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Green Gandalf
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Posted: 26th Aug 2014 16:22
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As HockeyKid says, you'll need l'Hopital's rule because the limit of both numerator and denominator is zero.

It's difficult to be more specific because we don't know whether you are supposed to know that while answering your question, or whether you're supposed to use formal rules of mathematical proof to determine the answer.



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Hockeykid
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Posted: 26th Aug 2014 19:25 Edited at: 26th Aug 2014 19:44
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Quote: "Practicing my LaTex on HockeyKid's reply"


That's pretty cool, but those multiplications (2, 1/2, 1/8, and 2/8) should all be inside of the limit, not outside the parenthesis.

Quote: "As HockeyKid says, you'll need l'Hopital's rule because the limit of both numerator and denominator is zero."


Without l'Hopital's rule it's possible using Lim x-> 0 sin(x)/ x = 1. However, I agree l'Hospital's rule is easier but I don't think he knows derivatives yet.


Sean

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Libervurto
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Posted: 26th Aug 2014 19:57
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@HockeyKid - Ah, yes I was thinking of integrals.

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Green Gandalf
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Posted: 28th Aug 2014 22:36
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Quote: "Without l'Hopital's rule it's possible using Lim x-> 0 sin(x)/ x = 1. "


Isn't that what you're asked to show - with different arguments for the sin()? If not then I agree. This is a good example where it helps to know what is being assumed as known.



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easter bunny
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Posted: 29th Aug 2014 06:12
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Thanks for the help guys.

I've got another problem, I need to differentiate x+3/x-2, but I'm not exactly sure how to do it
Do I do the top and bottom separately, or do I change it to (x+3)*((x-2)^-1), or what? I tried it a few ways and got it wrong each time


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Hockeykid
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Posted: 29th Aug 2014 06:24
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Quote: "Do I do the top and bottom separately, or do I change it to (x+3)*((x-2)^-1)"


Neither, you need to use the quotient shortcut (or you could use the definition of a derivative but that takes long).

(d/dx) (u/v) =

( v * (du/dx) - u * (dv/dx ) ) / v^2

So

( (x-2)(1) - (x+3)(1) ) / (x-2)^2

( x - 2 - x - 3 ) / (x-2)^2

-5/(x-2)^2 or -5/(x^2-4x+4)

Quote: "or do I change it to (x+3)*((x-2)^-1),"

This can only really be done in certain cases, usually when you just have a variable in the denominator or a variable times a constant in the denominator (like 1/x or 1/8x).


Sean

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easter bunny
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Posted: 29th Aug 2014 07:50 Edited at: 29th Aug 2014 07:51
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ohh, ok, thanks
Have this on the house


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Green Gandalf
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Posted: 29th Aug 2014 13:07
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Quote: "This can only really be done in certain cases, usually when you just have a variable in the denominator or a variable times a constant in the denominator (like 1/x or 1/8x)."


Not sure what you mean by that.

You can always do it that way if you have a ratio like u(x)/v(x). Just write it as u(x)*v(x)^(-1) and apply the usual rules for a product, etc. In fact in my nearly 50 years of doing calculus that's the way I've always done it - no need to remember another formula for division. The formula for differentiating a product is very easy to remember compared to the one for division.



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Hockeykid
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Posted: 29th Aug 2014 16:57
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Quote: "In fact in my nearly 50 years of doing calculus that's the way I've always done it "


Yeah I think I got that wrong, however I do believe in some cases the quotient rule can be easier than u(x)*v(x)^(-1) but either way will work.


Sean

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Green Gandalf
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Posted: 29th Aug 2014 21:50
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Quote: "but either way will work"


Agreed.

The downside to my version is that you generally need the function of a function rule as well - but that then makes two rules to recall rather than three. Still a bonus for those who prefer not to have to remember anything by rote.



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Seditious
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Posted: 29th Aug 2014 21:55
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If you haven't already, take a look at Khan Academy. It's a brilliant resource.

https://www.khanacademy.org/mission/differential-calculus
https://www.khanacademy.org/math/integral-calculus
https://www.khanacademy.org/math/multivariable-calculus
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Diggsey
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Posted: 30th Aug 2014 01:14
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Actually, you're over-complicating it. All you need to do is use the small angle approximation:

sin(x) -> x as x -> 0


In your examples, you can therefore replace sin(x) by x:

sin(2x)/8x = 2x/8x = 1/4
sin(9x)/6x = 9x/6x = 9/6
sin(-2x)/x = -2x/x = -2/1

[b]
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Green Gandalf
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Posted: 30th Aug 2014 12:39 Edited at: 30th Aug 2014 12:56
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Quote: "you can therefore replace sin(x) by x"


But then your first equality sign should be the approximately equal sign (which I can't find on my keyboard - one site says it's Alt+0247 but that is the division symbol on my laptop ). [Edit Found it! Had to go into Word first. ]

The important point is that the limits are as you stated. For any small non-zero x you don't have strict equality.

Edit So you could write



(which sadly doesn't actually tell you what the limit is - which is the whole point of the question I guess). Anyway, several people have answered this already.



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Clonkex
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Posted: 30th Aug 2014 15:27 Edited at: 30th Aug 2014 15:28
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Quote: "approximately equal sign (which I can't find on my keyboard"


I don't believe there is a key for it. The tilde key (left of the 1 key) can do the ~ symbol, which is the "approximately" symbol but not "approximately equal".

Quote: "one site says it's Alt+0247 but that is the division symbol on my laptop"


Yes, same for me too. However if you remove the 0 from the number and just do Alt+247 it works: ≈

EDIT: I find http://www.asciitable.com/ to be an excellent resource.


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Green Gandalf
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Posted: 30th Aug 2014 19:48
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Quote: "However if you remove the 0 from the number and just do Alt+247 it works"


Not for me.

In fact it has the bizarre effect as if I'd just deleted all this text and pressed my Browser's Home button.

Other times I get a weird small symbol depending on whether or not I've got Num Lock on.

Ah! Just twigged what's happening. With Num Lock off, the shortcut key for the Browser home page is Alt+Home (Home is the 7 key on the numeric key pad on this machine). With Num Lock on I get:¸ which looks a bit like a comma (,) but isn't. Weird.

Windows shortcuts to insanity if you ask me.



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easter bunny
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Posted: 1st Sep 2014 08:28
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Hit windows key + 'R', type charmap and look through the list until you find what you're looking for, double click it and use the copy bt\utton to copy it to the clipboard


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Green Gandalf
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Posted: 1st Sep 2014 13:14 Edited at: 1st Sep 2014 13:15
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Thanks. Didn't know about that.

Unfortunately it doesn't seem to work in any obvious way. Here's what I get when I try to copy the approximately equal symbol from the symbol sheet: »

Weird.

The character map screen looks OK as shown below. Any suggestions?





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Libervurto
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Posted: 4th Sep 2014 02:48
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Íñ łíñŭ× wé håvê æ çóm¶ößë kæý ſøŕ þåŧ.
Maybe there is something similar in windows?

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Clonkex
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Posted: 4th Sep 2014 03:29
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Quote: "Maybe there is something similar in windows?"


If only I could read what you'd written above this line....I don't know whether it was intentional, but this is what that line looks like on Windows:




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easter bunny
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Posted: 4th Sep 2014 03:57
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It says 'In linux we have a somethingoother key for bat'????
combo key for that?


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Clonkex
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Posted: 4th Sep 2014 04:12
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Quote: "combo key for that?"


Ah yes of course, that's what that says.


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Libervurto
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Posted: 4th Sep 2014 13:06 Edited at: 4th Sep 2014 13:08
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"In Linux we have a compose key for that." These are not b's: ß þ

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Clonkex
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Posted: 6th Sep 2014 05:39
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Quote: "These are not b's: ß þ"


No, ß is an S and þ is basically a Th sound. Groß means large or big in German and is pronounced "gross". þorn is the name of the letter þ and is pronounced (and spelled, in modern English) "thorn". I just couldn't read the words before and didn't look closely enough to work out what they were because I assumed they hadn't displayed properly.


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