Friday, 18 May 2018

calculus - Should L'Hopital's Rule be used for this limit?



Question



limx0(sinxx)11cosx



Attempt




This limit is equal to,



limx0exp(11cosxln(sinxx)).



I hope to solve this by focusing on the limit,



limx011cosxln(sinxx).



Which is indeterminate of the form "00". However, repeated application of L'Hopital's Rule seems to lead to an endless spiral of trigonometric terms, but wolfram Alpha says the limit I am focusing on exists.




Is there another way to solve this?


Answer



It should be used. There may be some little mistakes in your computation.



limx011cosxlnsinxx=limx0cosxsinx1xsinx=limx0xcosxsinxxsin2x=limx0xsinxsin2x+2xsinxcosx=limx01sinxx+2cosx=13


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