KevinSurrey
New member
- Oct 7, 2010
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Using Taylor series or otherwise, find the limit limx->0 sin 2x(cos(x)-1)/(1-e^(x^3))
Firstly i tried le hopitals rule as it is a 0/0 limit hoping i could find the answer that way. It is strange you then get 0/1 as x tends to 0. (The answer is 1)
I then used taylor polynomials with sin 2x being 2x-8/3!*x^3+2^5/5!*x^5 etc
and cos(x) 1-x^2/2+x^4/4! etc
and e^x^3 being 1+x^3+x^6/2!+x^9/3! etc
When you do this you end up with only terms in terms of variables of x left. I noticed that -x^3 appears on both top and bottom so suspect that something is meant to cancel to 1.
Please show me how to do this via taylor polynomials and why lehopitals does not work
also i realise I didnt attempt lehopitals correctly initially but I still cannot get the answer via lehopitals when i know i am doing it correctly
ok i got it with lehopitals so the answer is certainly 1 as it says in solutions-I look forward to the taylor series approach
Firstly i tried le hopitals rule as it is a 0/0 limit hoping i could find the answer that way. It is strange you then get 0/1 as x tends to 0. (The answer is 1)
I then used taylor polynomials with sin 2x being 2x-8/3!*x^3+2^5/5!*x^5 etc
and cos(x) 1-x^2/2+x^4/4! etc
and e^x^3 being 1+x^3+x^6/2!+x^9/3! etc
When you do this you end up with only terms in terms of variables of x left. I noticed that -x^3 appears on both top and bottom so suspect that something is meant to cancel to 1.
Please show me how to do this via taylor polynomials and why lehopitals does not work
also i realise I didnt attempt lehopitals correctly initially but I still cannot get the answer via lehopitals when i know i am doing it correctly
ok i got it with lehopitals so the answer is certainly 1 as it says in solutions-I look forward to the taylor series approach