# Limit of a function

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'Hopitals 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

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Char. Limit
Gold Member
Use L'Hopital's again, but first simplify the numerator.

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

alright so i applied l'hopitals 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
Gold Member
alright so i applied l'hopitals 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.

alright so i applied l'hopitals 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
Gold Member
Don't take the derivative again, this has a definite limit.
...no it doesn't...

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

...no it doesn't...

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Char. Limit
Gold Member
No problem. Just making sure the OP gets it right is all.