Find Limit of (x*cos x-sin x)/(x-sin⁡x) at x→0

[francisg3]

find the limit of lim ⁡(x*cos x-sin x)/(x-sin⁡x)
(x→0)



I know substitution does not work as it gives 0/0 and I attempted to factor and try the conjugate method without any result. I also tried l'hospital's rule which states that i take the derivative of the numerator and denominator which gave the following results:

lim x-> 0 = (cos x -xsin x-cos x)/(1-cos x) which still gives me 0/0

 

[Char. Limit]

Use L'Hopital's again, but first simplify the numerator.

 

[╔(σ_σ)╝]

You have to apply l'hospital's repeatedely for this guy or use the power series of sinx and cosx.

 

[francisg3]

alright so i applied l'hospital's again to
(-x*sin x)/(1-cos x) simplified version of the first derivative...then i dervied again to get

(-sin x - x*cos x)/(sin x) which itook the derivative once again to give:

(-cos x - cos x + x*sin x)/(cos x) which gives me a limit of -2...i have gone wrong somewhere

 

[Char. Limit]

alright so i applied l'hospital's again to
(-x*sin x)/(1-cos x) simplified version of the first derivative...then i dervied again to get

(-sin x - x*cos x)/(sin x) which itook the derivative once again to give:

(-cos x - cos x + x*sin x)/(cos x) which gives me a limit of -2...i have gone wrong somewhere

Don't know why you think that... -2 is the answer I get.

 

[Sourabh N]

alright so i applied l'hospital's again to
(-x*sin x)/(1-cos x) simplified version of the first derivative...then i dervied again to get

(-sin x - x*cos x)/(sin x) which itook the derivative once again to give:

Don't take the derivative again, this has a definite limit.

 

[Char. Limit]

Don't take the derivative again, this has a definite limit.

...no it doesn't...

 

[francisg3]

You are right Char. Limit, my mistake. Thanks for your help!

 

[Sourabh N]

...no it doesn't...

Eek I took x/Sin x = 1 in my head *headdesk*

 

[Char. Limit]

Eek I took x/Sin x = 1 in my head *headdesk*

No problem. Just making sure the OP gets it right is all.