Wednesday, 18 March 2015

limits - How to find limxto0fraccos(ax)cos(bx)cos(cx)sin(bx)sin(cx)



How to find limx0cos(ax)cos(bx)cos(cx)sin(bx)sin(cx)



I tried using L Hospital's rule but its not working!Help please!


Answer



Before using L'Hospital, turn the products to sums




\frac{\cos(ax)-\cos(bx)\cos(cx)}{\sin(bx)\sin(cx)}=\frac{2\cos(ax)-\cos((b-c)x)-\cos((b+c)x)}{\cos((b-c)x)-\cos((b+c)x)}.



Then by repeated application
\frac{2a\sin(ax)-(b-c)\sin((b-c)x)-(b+c)\sin((b+c)x)}{(b-c)\sin((b-c)x)-(b+c)\sin((b+c)x)},
and
\frac{2a^2\cos(ax)-(b-c)^2\cos((b-c)x)-(b+c)^2\cos((b+c)x)}{(b-c)^2\cos((b-c)x)-(b+c)^2\cos((b+c)x)}.



The limit is




\frac{2a^2-(b-c)^2-(b+c)^2}{(b-c)^2-(b+c)^2}=-\frac{a^2-b^2-c^2}{2bc}.


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