# Used l'hopitals rule too many times?

1. Apr 21, 2009

### a.a

1. The problem statement, all variables and given/known data
lim (sin x - x)/x as x--> 0

2. Relevant equations

3. The attempt at a solution

I used L'hopitals rule 3 times and got -1/6, but there is actually an asymptote at x = 0 so the limit should approach infinity or -infinity (in this case -infinity)
Where did I go wrong? Is it because I used l'hopitals rule too many times or inappropriately?

2. Apr 21, 2009

### dx

Re: Limit

Why did you use it three times? You just need to use it once.

3. Apr 21, 2009

### a.a

Re: Limit

but u still get 0/0 when u do this
sin x- x/x^3
=cos - 1 /3x^2
subs x=1
=0/0

4. Apr 21, 2009

### dx

Re: Limit

Not sure what you did there. If you differentiate the top and bottom, you get:

(cos(x) - 1)/1

5. Apr 21, 2009

### a.a

Re: Limit

sorry the original question:
lim (sin x - x)/x^3 as x-->0

6. Apr 21, 2009

### a.a

Re: Limit

so u get:
(cos(x) - 1)/(3x^2)

7. Apr 21, 2009

### dx

Re: Limit

Yes, in that case you use L'Hospital twice. And your answer of -1/6 is correct.

8. Apr 21, 2009

### a.a

Re: Limit

But if you take x = 0.0001 you get a very small -ve number
and take x = -0.0001 you get a small positive number, as if there is an asymptote....

9. Apr 21, 2009

### dx

Re: Limit

It should be negative for -0.0001 too.

10. Apr 21, 2009

### a.a

Re: Limit

so why do we get an actual number for the limit rather than -infinity?

11. Apr 21, 2009

### dx

Re: Limit

Why do you think it should be minus infinity? What expression did you get after applying L'Hospital's rule twice?

12. Apr 21, 2009

### a.a

Re: Limit

Because I was told that the graph has an asymptote at x=0
I was able to solve using L'hopitals rule and did not get an asymptote. That's why I'm confused...

13. Apr 21, 2009

### dx

Re: Limit

No there's no asymptote at x = 0. What exactly do you mean by asymptote?

14. Apr 21, 2009

### dx

15. Apr 21, 2009

### a.a

Re: Limit

Sorry, I guess the teacher was wrong, the function is defined at x=0 then.
Thanks

16. Apr 21, 2009

### Mentallic

Re: Limit

I'm unfamiliar with L'hospital's rules of differentiation, but:

d/dx((sinx-1)/x^3)=(xcosx-3sinx+3)/x^4

rather than what you both did which was differentiating the numerator and denominator separately.

17. Apr 21, 2009

### Matterwave

Re: Limit

L'hopital's rule has you differentiating the bottom and the top separately. It's not a rule of differentiation but a rule of limits, namely that if the limit of a quotient is in one of several forms (0/0 or such) then it is the same as the limit of the quotient of the derivatives. When they say "take the derivative" they mean "take the derivative of the top and then the bottom".

18. Apr 21, 2009

### Mentallic

Re: Limit

Ahh thankyou Matterwave.

19. Apr 22, 2009

### HallsofIvy

Re: Limit

I don't see why you would use L'Hopital at all.
$$lim_{x\rightarrow 0}\frac{sin x- x}{x}= \lim_{x\rightarrow 0}\frac{sin x}{x}- 1= 1- 1= 0$$

20. Apr 22, 2009

### dx

Re: Limit

Hi Halls,

He posted the question wrong, see post #5.