Find limit as x->0 of tan(ax)/sin(bx) please help :-D
R rmbiggee New member Joined Sep 11, 2008 Messages 1 Sep 11, 2008 #1 Find limit as x->0 of tan(ax)/sin(bx) please help :-D
skeeter Elite Member Joined Dec 15, 2005 Messages 3,218 Sep 11, 2008 #2 tan(ax)sin(bx)⋅abxabx=\displaystyle \frac{\tan(ax)}{\sin(bx)} \cdot \frac{abx}{abx} =sin(bx)tan(ax)⋅abxabx= tan(ax)abx⋅abxsin(bx)=\displaystyle \frac{\tan(ax)}{abx} \cdot \frac{abx}{\sin(bx)} =abxtan(ax)⋅sin(bx)abx= ab⋅tan(ax)ax⋅bxsin(bx)\displaystyle \frac{a}{b} \cdot \frac{\tan(ax)}{ax} \cdot \frac{bx}{\sin(bx)}ba⋅axtan(ax)⋅sin(bx)bx now take the limit as x -> 0.
tan(ax)sin(bx)⋅abxabx=\displaystyle \frac{\tan(ax)}{\sin(bx)} \cdot \frac{abx}{abx} =sin(bx)tan(ax)⋅abxabx= tan(ax)abx⋅abxsin(bx)=\displaystyle \frac{\tan(ax)}{abx} \cdot \frac{abx}{\sin(bx)} =abxtan(ax)⋅sin(bx)abx= ab⋅tan(ax)ax⋅bxsin(bx)\displaystyle \frac{a}{b} \cdot \frac{\tan(ax)}{ax} \cdot \frac{bx}{\sin(bx)}ba⋅axtan(ax)⋅sin(bx)bx now take the limit as x -> 0.
