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hba
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\[\lim_{x \rightarrow 0}\frac{ \sin2x }{ \sin3x }\]

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2/3

hba
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Please Explain :(

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ok

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one min

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since you have indeterminate form, you have to use L'hospital rule

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then when you do derivative you get \[\frac{ 2\cos2x }{ 3\cos3x }\]

hba
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I dont know l hospitals rule.

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you don't?

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you know like squeeze theorem?

hba
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Please teach me how to solve it,i am helpless :(

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oki so there are couple cases when you have indeterminate form, for example 0/0 , inf/inf

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so what you do then is basically take the derivative of the top and the bottom

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for example \[\lim_{x \rightarrow 1} \frac{ x^3+x^2=2x }{ x1 }\]

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it suppose to be x^3+x^22x

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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