# Used l'hopitals rule too many times?

## Homework Statement

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

## 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?

dx
Homework Helper
Gold Member

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

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
Homework Helper
Gold Member

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

(cos(x) - 1)/1

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

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

dx
Homework Helper
Gold Member

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

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
Homework Helper
Gold Member

It should be negative for -0.0001 too.

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

dx
Homework Helper
Gold Member

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

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...

dx
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Gold Member

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

dx
Homework Helper
Gold Member

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

Mentallic
Homework Helper

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
Gold Member

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
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Ahh thankyou Matterwave.

HallsofIvy
Homework Helper

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
Homework Helper
Gold Member

Hi Halls,

He posted the question wrong, see post #5.

HallsofIvy
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