lim (e^(ax)-e^(bx))/(sen(ax)-sen(bx)) = 0/0
x-->0
Aplique a regra de L'Hopital
lim (a*e^(ax)-b*e^(bx))/( a *cos(ax)- b*cos(bx))
= (a*e^(a*0)-b*e^(b*0))/( a *cos(a*0)- b*cos(b*0))
= (a*e^(0)-b*e^(0))/( a *cos(0)- b*cos(0))
= (a*1-b*1/( a *1- b1)
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lim (e^(ax)-e^(bx))/(sen(ax)-sen(bx)) = 0/0
x-->0
Aplique a regra de L'Hopital
lim (a*e^(ax)-b*e^(bx))/( a *cos(ax)- b*cos(bx))
x-->0
= (a*e^(a*0)-b*e^(b*0))/( a *cos(a*0)- b*cos(b*0))
= (a*e^(0)-b*e^(0))/( a *cos(0)- b*cos(0))
= (a*1-b*1/( a *1- b1)
=(a-b)/(a-b) = 1 ..para a≠b