This one sounds tough 6sin - 4h - 20h / sin 2h

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This one sounds tough 6sin - 4h - 20h / sin 2h

Mathematics
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what are we doing....limit, solving, simplifying ??
oh my bad limit
h --> 0

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Angles? 6sin? - 4h - 20h / sin 2h
like before take derivative of both top and bottom
okay i may sound like a complete re re, but the derivative is like f (a +h - f(a) / h or something?
what is the derivative?
yup, that is the definition of a derivative, there are some short forms to solve them though, which make life alot easier. for example the derivative of any function of the form x^n is n*x^(n-1) and such simple things like the derivative of sin(x) is cos(x), the derivative of cox(x) is -sin(x). if you haven't taken calculus yet though you probably shouldn't be using l'hopital's rule to solve these, just be using limit properties
so to solve this i take 6 sin - 4(a +h/h) - 20(a+h/h) / sin 2(a +h/h)?
oh right sorry if you haven't done derivatives yet. so you will have to try values really close to 0 on either side of 0 like -.001 and .001
you don't want to find the derivative that way for these functions, it will take too long and can become confusing
listen to dumbcow, you have a situation on the last one where you have 0 over 0, which normally we would use calculus to solve, but if you don't know it we must resort to other methods.
i'm in calculus, and she's foreign so i can barely understand her, and she doesnt explain how she got the answers, so in the case of the derivative, i have no idea what a or h is
one student said something about rolling the powers?
okay, well here's some rules for derivatives to skip that a and h stuff.\[y=x^n \rightarrow y'=nx^{n-1}\]\[y=sin(x)\rightarrow y'=cos(x)\]\[y=cos(x)\rightarrow y'=-sin(x)\]
so could you show me how that works in the equation?
hey dleather dont remember all of this but you should check out www.justmathtutoring.com free vids, walks you step by step on limits and everthing
i like asking on here so if i have any questions i can ask them, but i will be sure to go watch it though bc limits are giving me trouble
k, well when you run across the case of a limit where you either have 0 over 0 or infinity over 0 or 0 over infinity or infinity over infinity, you use what's known as l'hopital's rule, which states that the limit of a rational function is equal to the limit of the derivatives of top and bottom, mathematically that means this: \[lim~ x\rightarrow0\frac{f(x)}{g(x)}=lim~ x\rightarrow0\frac{f'(x)}{g'(x)}\]
so you just flip it over?
So in the case of your question you have a case where you have:\[lim ~x\rightarrow0~\frac{20h}{sin(2h)}\] so from that definition up there, you have f(h) = 20h and g(h) = sin(2h), following so far?
where does 6 sin - 4h go too?
we're just going to ignore it for a second, you can separate it out of the limit and do a direct substitution of 0 into it, which is pretty easy, we'll focus on the hard part of the question for now. I can include it all if you'd like, but its alot more typing.
whichever you would like to do
i just appreciate you helping me
okay, well let's deal with it now actually, in the original you can split the limit into two separate limits: \[lim~ h\rightarrow0~(6sin(-4h))-lim~ x\rightarrow0\frac{20h}{sin2h}\]the first limit just goes to zero when h goes to 0 so that's solved now.
i thinnnnk i see the answer is it 1/10?
and now we're left with just the second limit.\[lim~ h\rightarrow0\frac{20h}{sin(2x)}\]so using l'hopital's rule which I outlined above we take the derivative of the top and the derivative of the bottom: step by step is like this. \[f(h)=20h\rightarrow~f'(h)=20\]\[g(h)=sin(2h)\rightarrow~g'(h)=2cos(2h)\]so we place these back into the quotient: \[lim~ h\rightarrow0\frac{20}{2cos(2h)}\] where we can just use a direct substitution, and the limit is 10.
well -10 since theres that negative sign in the original expression
it still says its wrong i tried both 10 and -10
was it supposssed to be sin? instead of cos?
just making sure, the original expression looks like this:\[lim~ h\rightarrow0~~~ 6sin(-4h)-\frac{20h}{sin2h}\] right?
sin2h isunder the whole problem
okay what if i plug 0 in an equation and i come out with -1/ 0 then what do i use?
okay, well that changes things. still the same method though:\[ f(h)=6sin(-4h)-20h\rightarrow~f'(h)=-24cos(-4x)-20\]\[g'(h)=2cos(2x) ~~~\text{still}\] and now use a direct substitution.
should get -22
gotcha, so if i plug one in another equation and come out with -1/ 0 what does that mean, l hopitals rule?, and i already used my five trys on this homework problem, but you have helped me immensely regardless of if i got it right or not
okay, so if you have -1/0 you can't use l'hopital's rule, that's for the case of\[\frac{0~ or~ \infty}{0 ~or~ \infty}\] only.
so what do i do if it does come out to -1/0
you'll have to use limit properties, as a general rule those limits will either approach positive or negative infinity or just not exist, it depends on the functions involved though. I'm sorry, I can't stick around anymore though, I've got to go to bed, I'm sure Madcow will help you out though.
and by madcow i mean dumbcow
haha thanks man for your help! I gotta get going too, calculus at 8:15 in tha morning
best of luck!
same to you

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