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

• francisg3
In summary, the conversation revolved around finding the limit of a complex function using various methods such as substitution, factoring, and L'Hopital's rule. The final answer was determined to be -2, with some initial mistakes and misunderstandings along the way.

#### 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'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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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

francisg3 said:
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.

francisg3 said:
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.

Sourabh N said:
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!

Char. Limit said:
...no it doesn't...

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Sourabh N said:

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