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Mathematics 25 Online
OpenStudy (anonymous):

limit problem

OpenStudy (anonymous):

last one i need help on (: @aegidious

OpenStudy (anonymous):

OpenStudy (anonymous):

Already here :)

OpenStudy (anonymous):

oooooooh okay, this one is going to deal with trigonometric identities. Do you remember how we stretched out the numerator of the last equation with what it equaled? Well we're going to do the same thing here with trigonometric identities that equal sine so we can stretch out the denominator into something else

OpenStudy (anonymous):

ohhhh

OpenStudy (anonymous):

mmmmhmmm aren't trig identities fun? -_- *sarcasm* lol

OpenStudy (anonymous):

okay ummm how to start...

OpenStudy (anonymous):

hahaha oh yeahhh

OpenStudy (anonymous):

i did it before and got 2/5 and i was going by my textbook but i dont think its right

OpenStudy (anonymous):

well you understand that tan(x) = sin(x) / cos(x) right? then that must also mean that sin(x) = cos(x)*tan(x)

OpenStudy (anonymous):

yess

OpenStudy (anonymous):

hmmm....

OpenStudy (anonymous):

trig problems make me nervous.... give me a minute to think...

OpenStudy (anonymous):

hahaha okay

OpenStudy (anonymous):

hmmmm does your textbook give you 2/5? or was that what you got?

OpenStudy (anonymous):

thats what i got

OpenStudy (anonymous):

it could be totally wrong

OpenStudy (anonymous):

okay, now say we have used trig identities to get tan(-2x)*cos(-2x) --------------- tan(-5x)*cos(-5x) we need to use trig identities to get the tan out of it. 2/5 would be right if the tan wasn't there, but it's just making the whole thing complicated -_- so hmmmm

OpenStudy (anonymous):

and you can't simply factor the tangents out, because the x's are being multiplied by two different constants...

OpenStudy (anonymous):

oh this is really hard hahah

OpenStudy (anonymous):

mmmmhmmmmm tell me about it...

OpenStudy (anonymous):

ughhh nobody wants to help

OpenStudy (anonymous):

i put it into wolfram and got 2/5 (:

OpenStudy (anonymous):

Using the fact that sine is an odd function, we can reduce this (somewhat) to \[\frac{\sin(-2x)}{\sin(-5x)}=\frac{-\sin2x}{-\sin5x}=\frac{\sin2x}{\sin5x}\] Now try rewriting the expression so that you can make use of the known limits, \[\lim_{x\to0}\frac{\sin ax}{ax}=\lim_{x\to0}\frac{ax}{\sin ax}=1\] for \(a\not=0\).

OpenStudy (anonymous):

you got 1? @SithsAndGiggles

OpenStudy (anonymous):

@jim_thompson5910 what do you get?

OpenStudy (anonymous):

I said no such thing. I'm giving you hints to get the right answer, which is indeed 2/5.

jimthompson5910 (jim_thompson5910):

are you allowed to use L'Hospital's Rule?

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