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anonymous
 one year ago
Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it.
lim x→0
e^2x − e^−2x − 4x/xsinx
I've tried to do it w the rule but i keep getting it wrong, please help!!
anonymous
 one year ago
Find the limit. Use l'Hospital's Rule if appropriate. If there is a more elementary method, consider using it. lim x→0 e^2x − e^−2x − 4x/xsinx I've tried to do it w the rule but i keep getting it wrong, please help!!

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caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0\[\lim_{x \rightarrow 0}e ^{2x}e ^{2x}\frac{ 4x }{ x }\sin(x)\]

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0am I correct in transcribing your question?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0almost, the whole thing is being divided by xsinx

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0For starters, is using the rule appropriate?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well i plugged 0 in and i ended up with 0/0

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0@SithsAndGiggles I learned this rule in calc 1 so I guess yes?

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0@gaba a joke first: does you name means the neurotransmitter gammaaminobutyic acid?

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0I see what you mean so the correct version would be \[\lim_{x \rightarrow 0}\frac{ e^{2x}e ^{2x}4x }{ x\sin(x) }\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0gaba its one of my nicknames, but yeah jaja

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0Do you now what exactly is lhospital rule and how to use it?

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0If you don't then that's the end of game, but if you do it's a piece of cake

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sort of, but i keep getting the same thing over and over again, so im confused

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0Ok, so you know the new limit is formed by taking the quotient of the derivatives of the numerator and the denominator, right?

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0so what are the derivatives we need?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0e^2xe^2x4/xcosx?

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0OK, I have to say you kind of don't know your stuff

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{ de ^{nx} }{ dx }= n*e^{nx}\]

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0n is integer, for now

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0you still there? what would be the correct answer?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i got 2e^2x+2e^2x4 / 1cosx

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0but it is still 0 over 0 right?

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0so you got to do one more round

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yeah, and i did but i kept getting that

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0so 4e^2x+4e^2x/sin(x)

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0wait.... that would be 8/0....WTF

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0my brain is so rusty cuz I had my calc 1 3 years ago

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0right now i have 16e^2x16e^2x / sinx

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0how do you get your answer?! I didn't get any 16 or sin

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What I meant by my comment was, does the limit satisfy the conditions for using L'Hopital's rule? The answer is yes, because substituting \(x=0\) directly gives you a proper indeterminate form: \[\frac{e^{2(0)}e^{2(0)}4(0)}{0\sin 0}=\frac{e^0e^00}{00}=\frac{0}{0}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0i kept going, i had 2e^2x, and then 4e^2x, and then 8e^2x because i kept getting 0/0 with all of them

caozeyuan
 one year ago
Best ResponseYou've already chosen the best response.0But I stoped on the second trial cuz the upstair is 8

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\begin{align*}\lim_{x\to0}\frac{e^{2x}e^{2x}4x}{x\sin x}&=\lim_{x\to0}\frac{\dfrac{d}{dx}\left[e^{2x}e^{2x}4x\right]}{\dfrac{d}{dx}\left[x\sin x\right]}\\\\ &=\lim_{x\to0}\frac{2e^{2x}+2e^{2x}4}{1\cos x}=\frac{2e^0+2e^04}{1\cos0}=\frac{0}{0}\\\\ &=\lim_{x\to0}\frac{\dfrac{d}{dx}\left[2e^{2x}+2e^{2x}4\right]}{\dfrac{d}{dx}\left[1\cos x\right]}\\\\ &=\lim_{x\to0}\frac{4e^{2x}4e^{2x}}{\sin x}=\frac{4e^04e^0}{\sin0}=\frac{0}{0}\\\\ &=\lim_{x\to0}\frac{\dfrac{d}{dx}\left[4e^{2x}4e^{2x}\right]}{\dfrac{d}{dx}[\sin x]}\\\\ &=\lim_{x\to0}\frac{8e^{2x}+8e^{2x}}{\cos x}=\frac{8e^0+8e^0}{\cos0}\neq\frac{0}{0}\end{align*}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so then my answer is 16?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sorry, I meant, not negative.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Yes, just to check: http://www.wolframalpha.com/input/?i=Limit%5B%282+Sinh%5B2x%5D4x%29%2F%28xSin%5Bx%5D%29%2Cx%3E0%5D
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