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

- anonymous

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

- Stacey Warren - Expert brainly.com

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- schrodinger

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- dumbcow

what are we doing....limit, solving, simplifying ??

- anonymous

oh my bad limit

- anonymous

h --> 0

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## More answers

- Directrix

Angles?
6sin? - 4h - 20h / sin 2h

- dumbcow

like before take derivative of both top and bottom

- anonymous

okay i may sound like a complete re re, but the derivative is like f (a +h - f(a) / h or something?

- anonymous

what is the derivative?

- anonymous

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

- anonymous

so to solve this i take 6 sin - 4(a +h/h) - 20(a+h/h) / sin 2(a +h/h)?

- dumbcow

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

- dumbcow

you don't want to find the derivative that way for these functions, it will take too long and can become confusing

- anonymous

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.

- anonymous

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

- anonymous

one student said something about rolling the powers?

- anonymous

okay, well here's some rules for derivatives to skip that a and h stuff.\[y=x^n \rightarrow y'=nx^{n-1}\]\[y=sin(x)\rightarrow y'=cos(x)\]\[y=cos(x)\rightarrow y'=-sin(x)\]

- anonymous

so could you show me how that works in the equation?

- anonymous

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

- anonymous

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

- anonymous

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: \[lim~ x\rightarrow0\frac{f(x)}{g(x)}=lim~ x\rightarrow0\frac{f'(x)}{g'(x)}\]

- anonymous

so you just flip it over?

- anonymous

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?

- anonymous

where does 6 sin - 4h go too?

- anonymous

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.

- anonymous

whichever you would like to do

- anonymous

i just appreciate you helping me

- anonymous

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.

- anonymous

i thinnnnk i see the answer is it 1/10?

- anonymous

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.

- anonymous

well -10 since theres that negative sign in the original expression

- anonymous

it still says its wrong i tried both 10 and -10

- anonymous

was it supposssed to be sin? instead of cos?

- anonymous

just making sure, the original expression looks like this:\[lim~ h\rightarrow0~~~ 6sin(-4h)-\frac{20h}{sin2h}\] right?

- anonymous

sin2h isunder the whole problem

- anonymous

okay what if i plug 0 in an equation and i come out with -1/ 0 then what do i use?

- anonymous

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.

- anonymous

should get -22

- anonymous

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

- anonymous

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.

- anonymous

so what do i do if it does come out to -1/0

- anonymous

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.

- anonymous

and by madcow i mean dumbcow

- anonymous

haha thanks man for your help! I gotta get going too, calculus at 8:15 in tha morning

- anonymous

best of luck!

- anonymous

same to you

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