Question

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!!

Answer 1

\[\lim_{x \rightarrow 0}e ^{2x}-e ^{-2x}-\frac{ 4x }{ x }-\sin(x)\]

Answer 2

am I correct in transcribing your question?

Answer 3

almost, the whole thing is being divided by x-sinx

Answer 4

For starters, is using the rule appropriate?

Answer 5

well i plugged 0 in and i ended up with 0/0

Answer 6

@SithsAndGiggles I learned this rule in calc 1 so I guess yes?

Answer 7

@gaba a joke first: does you name means the neurotransmitter gamma-amino-butyic acid?

Answer 8

I see what you mean so the correct version would be \[\lim_{x \rightarrow 0}\frac{ e^{2x}-e ^{-2x}-4x }{ x-\sin(x) }\]

Answer 9

gaba its one of my nicknames, but yeah jaja

Answer 10

yeah that's it

Answer 11

Do you now what exactly is lhospital rule and how to use it?

Answer 12

If you don't then that's the end of game, but if you do it's a piece of cake

Answer 13

sort of, but i keep getting the same thing over and over again, so im confused

Answer 14

Ok, so you know the new limit is formed by taking the quotient of the derivatives of the numerator and the denominator, right?

Answer 15

yes

Answer 16

so what are the derivatives we need?

Answer 17

e^2x-e^-2x-4/x-cosx?

Answer 18

OK, I have to say you kind of don't know your stuff

Answer 19

\[\frac{ de ^{nx} }{ dx }= n*e^{nx}\]

Answer 20

n is integer, for now

Answer 21

you still there? what would be the correct answer?

Answer 22

i got 2e^2x+2e^-2x-4 / 1-cosx

Answer 23

but it is still 0 over 0 right?

Answer 24

so you got to do one more round

Answer 25

yeah, and i did but i kept getting that

Answer 26

so 4e^2x+4e^2x/sin(x)

Answer 27

wait.... that would be 8/0....WTF

Answer 28

my brain is so rusty cuz I had my calc 1 3 years ago

Answer 29

right now i have 16e^2x-16e^-2x / -sinx

Answer 30

undefined right?

Answer 31

how do you get your answer?! I didn't get any 16 or -sin

Answer 32

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:
\[\frac{e^{2(0)}-e^{-2(0)}-4(0)}{0-\sin 0}=\frac{e^0-e^0-0}{0-0}=\frac{0}{0}\]

Answer 33

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

Answer 34

But I stoped on the second trial cuz the upstair is 8

Answer 35

\[\begin{align*}\lim_{x\to0}\frac{e^{2x}-e^{-2x}-4x}{x-\sin x}&=\lim_{x\to0}\frac{\dfrac{d}{dx}\left[e^{2x}-e^{-2x}-4x\right]}{\dfrac{d}{dx}\left[x-\sin x\right]}\\\\
&=\lim_{x\to0}\frac{2e^{2x}+2e^{-2x}-4}{1-\cos x}=\frac{2e^0+2e^0-4}{1-\cos0}=\frac{0}{0}\\\\
&=\lim_{x\to0}\frac{\dfrac{d}{dx}\left[2e^{2x}+2e^{-2x}-4\right]}{\dfrac{d}{dx}\left[1-\cos x\right]}\\\\
&=\lim_{x\to0}\frac{4e^{2x}-4e^{-2x}}{\sin x}=\frac{4e^0-4e^0}{\sin0}=\frac{0}{0}\\\\
&=\lim_{x\to0}\frac{\dfrac{d}{dx}\left[4e^{2x}-4e^{-2x}\right]}{\dfrac{d}{dx}[\sin x]}\\\\
&=\lim_{x\to0}\frac{8e^{2x}+8e^{-2x}}{-\cos x}=\frac{8e^0+8e^0}{\cos0}\neq\frac{0}{0}\end{align*}\]

Answer 36

so then my answer is -16?

Answer 37

Sorry, I meant, not negative.

Answer 38

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

Answer 39

thank you so much!!

Answer 40

yw