Used l'hopitals rule too many times?

[a.a]

Homework Statement

lim (sin x - x)/x as x--> 0

Homework Equations

The Attempt at a Solution

I used l'hospital's 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'hospital's rule too many times or inappropriately?

 

[dx]

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

 

[a.a]

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

 

[dx]

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

(cos(x) - 1)/1

 

[a.a]

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

 

[a.a]

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

 

[dx]

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

 

[a.a]

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...

 

[dx]

It should be negative for -0.0001 too.

 

[a.a]

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

 

[dx]

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

 

[a.a]

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

 

[dx]

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

 

[dx]

Here's a plot:

http://img254.imagevenue.com/img.php?image=44618_plot_122_93lo.jpg

 

[a.a]

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

 

[Mentallic]

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.

 

[Matterwave]

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.

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".

 

[Mentallic]

Ahh thankyou Matterwave.

 

[HallsofIvy]

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

 

[dx]

Hi Halls,

He posted the question wrong, see post #5.

 

[HallsofIvy]

Thanks, that's what I get for not reading the whole thread!

 

[statdad]

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.

L'hospital's theorem is a method of evaluating limits in certain situations. The procedure doesn't require the usual rule for differentiating a quotient: the correct approach (in this particular problem) actually does call for differentiating the numerator and denominator separately: the work as shown in prior posts is correct.