I was working on this question.
limx→0asinbx−bsinaxx2sinax
limx→01x2⋅limx→01abx1abx⋅asinbx−bsinaxsinax
limx→01x2⋅limx→0sinbxbx−limx→0sinaxax1blimx→0sinaxax
blimx→01x2⋅b−aa
It seems like this limit does not exist, but if you apply L'Hopital's rule you seem to get an answer. What is wrong with what I did?
Answer
One error is that
limx→0sincxcx=1, not c,
so your fraction is
1x21−11/b
which is still indeterminate.
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