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

1. caozeyuan

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

2. caozeyuan

am I correct in transcribing your question?

3. anonymous

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

4. anonymous

For starters, is using the rule appropriate?

5. anonymous

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

6. caozeyuan

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

7. caozeyuan

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

8. caozeyuan

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

9. anonymous

gaba its one of my nicknames, but yeah jaja

10. anonymous

yeah that's it

11. caozeyuan

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

12. caozeyuan

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

13. anonymous

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

14. caozeyuan

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

15. anonymous

yes

16. caozeyuan

so what are the derivatives we need?

17. anonymous

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

18. caozeyuan

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

19. caozeyuan

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

20. caozeyuan

n is integer, for now

21. caozeyuan

you still there? what would be the correct answer?

22. anonymous

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

23. caozeyuan

but it is still 0 over 0 right?

24. caozeyuan

so you got to do one more round

25. anonymous

yeah, and i did but i kept getting that

26. caozeyuan

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

27. caozeyuan

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

28. caozeyuan

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

29. anonymous

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

30. anonymous

undefined right?

31. caozeyuan

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

32. anonymous

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

33. anonymous

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

34. caozeyuan

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

35. anonymous

\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*}

36. anonymous

so then my answer is -16?

37. anonymous

Sorry, I meant, not negative.

38. anonymous
39. anonymous

thank you so much!!

40. anonymous

yw