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  • hba
\[\lim_{x \rightarrow 0}\frac{ \sin2x }{ \sin3x }\]
  • hba
Please Explain :(

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Other answers:

one min
since you have indeterminate form, you have to use L'hospital rule
then when you do derivative you get \[\frac{ 2\cos2x }{ 3\cos3x }\]
  • hba
I dont know l hospitals rule.
you don't?
you know like squeeze theorem?
  • hba
Please teach me how to solve it,i am helpless :(
oki so there are couple cases when you have indeterminate form, for example 0/0 , inf/inf
so what you do then is basically take the derivative of the top and the bottom
for example \[\lim_{x \rightarrow 1} \frac{ x^3+x^2=2x }{ x-1 }\]
it suppose to be x^3+x^2-2x
so if you plug 1 in you get 0/0 which is indeterminate as it has no meaning. so you have to apply l'hospital, derivative of the top and bottom

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