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/x-sinx I've tried to do it w the rule but i keep getting it wrong, please help!!

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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/x-sinx I've tried to do it w the rule but i keep getting it wrong, please help!!

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\[\lim_{x \rightarrow 0}e ^{2x}-e ^{-2x}-\frac{ 4x }{ x }-\sin(x)\]
am I correct in transcribing your question?
almost, the whole thing is being divided by x-sinx

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For starters, is using the rule appropriate?
well i plugged 0 in and i ended up with 0/0
@SithsAndGiggles I learned this rule in calc 1 so I guess yes?
@gaba a joke first: does you name means the neurotransmitter gamma-amino-butyic acid?
I see what you mean so the correct version would be \[\lim_{x \rightarrow 0}\frac{ e^{2x}-e ^{-2x}-4x }{ x-\sin(x) }\]
gaba its one of my nicknames, but yeah jaja
yeah that's it
Do you now what exactly is lhospital rule and how to use it?
If you don't then that's the end of game, but if you do it's a piece of cake
sort of, but i keep getting the same thing over and over again, so im confused
Ok, so you know the new limit is formed by taking the quotient of the derivatives of the numerator and the denominator, right?
yes
so what are the derivatives we need?
e^2x-e^-2x-4/x-cosx?
OK, I have to say you kind of don't know your stuff
\[\frac{ de ^{nx} }{ dx }= n*e^{nx}\]
n is integer, for now
you still there? what would be the correct answer?
i got 2e^2x+2e^-2x-4 / 1-cosx
but it is still 0 over 0 right?
so you got to do one more round
yeah, and i did but i kept getting that
so 4e^2x+4e^2x/sin(x)
wait.... that would be 8/0....WTF
my brain is so rusty cuz I had my calc 1 3 years ago
right now i have 16e^2x-16e^-2x / -sinx
undefined right?
how do you get your answer?! I didn't get any 16 or -sin
What 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: \[\frac{e^{2(0)}-e^{-2(0)}-4(0)}{0-\sin 0}=\frac{e^0-e^0-0}{0-0}=\frac{0}{0}\]
i kept going, i had 2e^2x, and then 4e^2x, and then 8e^2x because i kept getting 0/0 with all of them
But I stoped on the second trial cuz the upstair is 8
\[\begin{align*}\lim_{x\to0}\frac{e^{2x}-e^{-2x}-4x}{x-\sin x}&=\lim_{x\to0}\frac{\dfrac{d}{dx}\left[e^{2x}-e^{-2x}-4x\right]}{\dfrac{d}{dx}\left[x-\sin x\right]}\\\\ &=\lim_{x\to0}\frac{2e^{2x}+2e^{-2x}-4}{1-\cos x}=\frac{2e^0+2e^0-4}{1-\cos0}=\frac{0}{0}\\\\ &=\lim_{x\to0}\frac{\dfrac{d}{dx}\left[2e^{2x}+2e^{-2x}-4\right]}{\dfrac{d}{dx}\left[1-\cos x\right]}\\\\ &=\lim_{x\to0}\frac{4e^{2x}-4e^{-2x}}{\sin x}=\frac{4e^0-4e^0}{\sin0}=\frac{0}{0}\\\\ &=\lim_{x\to0}\frac{\dfrac{d}{dx}\left[4e^{2x}-4e^{-2x}\right]}{\dfrac{d}{dx}[\sin x]}\\\\ &=\lim_{x\to0}\frac{8e^{2x}+8e^{-2x}}{-\cos x}=\frac{8e^0+8e^0}{\cos0}\neq\frac{0}{0}\end{align*}\]
so then my answer is -16?
Sorry, I meant, not negative.
Yes, just to check: http://www.wolframalpha.com/input/?i=Limit%5B%282+Sinh%5B2x%5D-4x%29%2F%28x-Sin%5Bx%5D%29%2Cx-%3E0%5D
thank you so much!!
yw

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