limit problem
last one i need help on (: @aegidious
Already here :)
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
ohhhh
mmmmhmmm aren't trig identities fun? -_- *sarcasm* lol
okay ummm how to start...
hahaha oh yeahhh
i did it before and got 2/5 and i was going by my textbook but i dont think its right
well you understand that tan(x) = sin(x) / cos(x) right? then that must also mean that sin(x) = cos(x)*tan(x)
yess
hmmm....
trig problems make me nervous.... give me a minute to think...
hahaha okay
hmmmm does your textbook give you 2/5? or was that what you got?
thats what i got
it could be totally wrong
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
and you can't simply factor the tangents out, because the x's are being multiplied by two different constants...
oh this is really hard hahah
mmmmhmmmmm tell me about it...
ughhh nobody wants to help
i put it into wolfram and got 2/5 (:
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 \(a\not=0\).
you got 1? @SithsAndGiggles
@jim_thompson5910 what do you get?
I said no such thing. I'm giving you hints to get the right answer, which is indeed 2/5.
are you allowed to use L'Hospital's Rule?
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