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lim x-->0 f(x) = (e^x - 1)/(sin x) = 0/0 L'hopital's rule
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\[f(x) =\frac{ (e^x-1)' }{ (sinx)' }\]
l'hospital rule states that if you have a lim that's 0/0 you can use f'x/g'x to get the limit
l'hospital rule states that if you have a lim that's 0/0 you can use f'x/g'x to get the limit
\[\frac{ e^x }{ \cos~x } = \frac{ e^0 }{ \cos 0 } = 1\] right? @Luigi0210
1/1=1 is right
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That's right
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