Question
limx→0(sinxx)11−cosx
Attempt
This limit is equal to,
limx→0exp(11−cosxln(sinxx)).
I hope to solve this by focusing on the limit,
limx→011−cosxln(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.
limx→011−cosxlnsinxx=limx→0cosxsinx−1xsinx=limx→0xcosx−sinxxsin2x=limx→0−xsinxsin2x+2xsinxcosx=−limx→01sinxx+2cosx=−13
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